APPENDIX A: FORMAL PROOFS

First I prove Lemmas 1–5, then characterize the PBE of interest (Lemmas 6–8), and finally prove Propositions 1–3 from the text.

Lemma 1. If Δ₁ = 1, G upholds $\left( {\phi _1^* = 1} \right)$.

Proof. Note that G never incurs κ _C from upholding when Δ₁ = 1. Thus, when G upholds, its expected utility is ${\mathbb{E}}\;U_G^1 = \chi \,\equiv \Pr \left[ {{x_C} = 1{\rm{|}}{x_G} = 1} \right] \,\ge\, 0$. When G vetoes, ${\mathbb{E}}\;U_G^0 = - {\kappa _A} \,\lt\, 0$. ■

Lemma 2. When C is informed of Δ₁ through review, C upholds with probability

(11)$${\chi ^{* + }} = \left\{ {\matrix{ 1 \hfill & {{\rm{if}}\;{\Delta _1} = 1} \hfill \cr 0 \hfill & {{\rm{if}}\;{\Delta _1} = - 1} \hfill \cr} \vskip6.3pt.} \right.$$

When C is uninformed of Δ₁, C upholds with probability

(12)$${\chi ^{* - }} = \left\{ {\matrix{ 1 \hfill & {{\rm{if}}\;{\phi _{ - 1}} \,\le\, \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)\;{\rm{or}}\;{\Delta _2} = 1} \hfill \cr 0 \hfill & {{\rm{if}}\;{\phi _{ - 1}} > \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)\;{\rm{and}}\;{\Delta _2} = - 1} \hfill \cr} \vskip9.5pt.} \right$$

Proof. In each case, C’s utility from x _C = 0 is 0.

1. 1. If C observes Δ₁ = 1, then its utility from x _C = 1 is at least (1 − σ) > 0, so C upholds. If C observes Δ₁ = −1, then its utility from upholding is at most −1, so C reverses.

2. 2. If C is uncertain of Δ₁, it can reverse only if Δ₂ = −1, in which case its expected utility of upholding is E[Δ₁|x _G = 1] − σ = 2d − 1 − σ. C’s expected utility from reversal is 0. Inserting d from equation (1), 2d − 1 − σ > 0 is equivalent to ${\phi _{ - 1}} \,\le\, \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)$. ■

Lemma 3. If γ > 0 and ϕ ₋₁ = 0, then $r_{{\Delta _2}}^* = 0$.

Proof. If Δ₂ = 1, equation (5) indicates that r* = 1 only if ${\phi _{ - 1}} \ge \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right)$. If 0 < γ < λ, then ${\phi _{ - 1}} = 0 \,\lt\, \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right)$. If γ ≥ λ, r* = 0 is a dominant strategy for C.

If Δ₂ = −1, equation (6) indicates that r* = 1 only if ${\phi _{ - 1}} \ge \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)$. If 0 < γ < λ(1 + σ), then ${\phi _{ - 1}} = 0 \,\lt\, \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)$. If γ ≥ λ(1 + σ), r* = 0 is a dominant strategy for C. ■

Lemma 4. If ϕ ₋₁ = 1 and

(13)$$\gamma > \left\{ {\matrix{ {\lambda \left( {1 - \delta } \right)} \hfill &#x0026; {{\rm{if}}\;{\Delta _2} = 1} \hfill \cr {\lambda \left( {1 - \delta } \right)\left( {1 + \sigma } \right)} \hfill &#x0026; {{\rm{if}}\;{\Delta _2} = - 1\;{\rm{and}}\;{\chi ^{* - }} = 1} \hfill \cr {\lambda \delta \left( {1 - \sigma } \right)} \hfill &#x0026; {{\rm{if}}\;{\Delta _2} = - 1\;{\rm{and}}\;{\chi ^{* - }} = 0} \hfill \cr} \vskip11pt,} \right.$$

then $r_{{\Delta _2}}^* = 0$.

Proof.

1. 1. If Δ₂ = 1, equation (5) indicates that r* = 1 only if γ ≤ λ(1 − d). When ϕ ₋₁ = 1, d = δ, so r* = 1 only if γ ≤ λ(1 − δ).

2. 2. If Δ₂ = −1, note that χ*⁻ = 1 and ϕ ₋₁ = 1 together imply σ ≤ 2δ − 1. Thus, equation (6) governs C’s best response, and indicates that r* = 1 only if γ ≤ λ(1 − d)(1 + σ). When ϕ ₋₁ = 1, this simplifies to γ ≤ λ(1 − δ)(1 + σ).

3. 3. If Δ₂ = −1 and χ*⁻ = 0, equation (7) governs C’s best response, and indicates that r* = 1 only if γ ≤ λd(1 − σ). When ϕ ₋₁ = 1, this simplifies to γ ≤ λδ(1 − σ). Note that λδ(1 − σ) < λ(1 − δ)(1 + σ) if and only if σ > 2δ − 1. ■

Lemma 5. If κ _A > 1 and ρ = 0, then $\phi _{ - 1}^* = 1$.

Proof. From equation (9), ${\mathbb{E}}\;U_G^1 = - \chi - \rho \lambda {\kappa _C}$ from approval when Δ₁ = −1, and ${\mathbb{E}}\;U_G^0 = - {\kappa _A}$ from veto when Δ₁ = −1. Thus κ _A > 1 and ρ = 0 imply ${\mathbb{E}}\;U_G^1 = - \chi \ge - 1 > - {\kappa _A} = {\mathbb{E}}\;U_G^0$. ■

The following results present the most informative PBE for each parameter configuration. Define (cf. Proposition 1)

$$\hat{\gamma }\left( {{\Delta _2},\sigma } \right) \equiv \left\{ {\matrix{ {\lambda \left( {1 - \delta } \right)} \hfill &#x0026; {{\rm{for}}\;{\Delta _2} = 1} \hfill \cr {\lambda \left( {1 - \delta } \right)\left( {1 + \sigma } \right)} \hfill &#x0026; {{\rm{for}}\;{\Delta _2} = - 1\;{\rm{and}}\;\sigma \,\le\, 2\delta - 1} \hfill \cr {\lambda \delta \left( {1 - \sigma } \right)} \hfill &#x0026; {{\rm{for}}\;{\Delta _2} = - 1\;{\rm{and}}\;\sigma > 2\delta - 1} \hfill \cr} \vskip12pt.} \right.$$

Lemma 6. Suppose Δ₂ = −1 and σ > 2δ − 1.

1. 1. If $\gamma \,\le\, \lambda \delta \left( {1 - \sigma } \right) \equiv \hat{\gamma }\left( {{\Delta _2} = - 1,\sigma > 2\delta - 1} \right)$, then in the most informative PBE, $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \right)$ and ${\rho ^*} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$. C reverses a law approved by G if review is informative and Δ₁ = −1; C upholds a law approved by G otherwise.

2. 2. If $\gamma > \hat{\gamma }\left( {{\Delta _2} = - 1,\sigma > 2\delta - 1} \right)$, then in the most informative PBE, $\left( {\phi _1^{\rm{*}},\phi _{ - 1}^{\rm{*}}} \right) = \left( {1,1} \right)$ and ρ* = 0. C reverses all laws approved by G.

Proof. First note that σ > 2δ − 1 and γ ≤ λδ(1 − σ) jointly imply $\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right) \le \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)$, so equation (6) is the relevant best response function. Second, σ > 2δ − 1 implies λδ(1 − σ) < λ(1 − δ)(1 + σ), where the latter is the largest γ such that C reviews under Lemma 4, so C is willing to review for γ ≤ λδ(1 − σ) and ${\phi _{ - 1}} \lt \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)$. Thus, if γ ≤ λδ(1 − σ), σ > 2δ − 1, and ${\phi _{ - 1}} = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)$, then by equation (6), C is indifferent between review and no review, so ${\rho ^*} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$ is a best response. If C reviews with probability $\rho = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$, then by equation (9), G is indifferent between upholding and vetoing when Δ₁ = −1, so $\phi _{ - 1}^* = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)$ is a best response.

Second, if σ > 2δ − 1, then ϕ ₋₁ = 1 ⇒ χ*⁻ = 0 by Lemma 2, so γ > λδ(1 − σ) implies ρ* = 0 by Lemma 4, and thus $\phi _{ - 1}^* = 1$ by Lemma 5.

For C’s approval, note γ ≤ λδ(1 − σ) ⇒ χ*⁻ = 1, and γ > λδ(1 − σ) ⇒ χ*⁻ = 0. Then x _C in each case follow from Lemma 2. ■

Lemma 7. Suppose Δ₂ = −1 and σ ≤ 2δ − 1.

1. 1. If $\gamma \,\le\, \lambda (1 - \delta )(1 + \sigma ) \equiv \hat{\gamma }({{\rm{\Delta }}_2} = - 1,\sigma \,\le\, 2\delta - 1)$, then in the most informative PBE, $\left( {\phi _1^{\rm{*}},\phi _{ - 1}^{\rm{*}}} \right) = \left( {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \right)$ and ${\rho ^{\rm{*}}} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$. C reverses a law approved by G if review is informative and Δ₁ = −1; C upholds a law approved by G otherwise.

2. 2. If $\gamma > \hat{\gamma }({\Delta _2} = - 1,\sigma \,\le\, 2\delta - 1)$, then in the most informative PBE, $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {1,1} \right)$ and ρ* = 0. C approves all laws approved by G.

Proof. First, if Δ₂ = −1 and σ ≤ 2δ − 1, then $\left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right) \ge 1$ and only equation (6) governs C’s best response. Thus, if γ ≤ λ(1 − δ)(1 + σ) and ${\phi _{ - 1}} = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)$, then by equation (6), C is indifferent between review and no review, so ${\rho ^*} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$ is a best response. If C reviews with probability $\rho = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$, then by equation (9), G is indifferent between upholding and vetoing when Δ₁ = −1, so $\phi _{ - 1}^* = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)$ is a best response.

Second, if σ ≤ 2δ − 1, then ϕ ₋₁ = 1 ⇒ χ*⁻ = 1 by Lemma 2, so γ > λ(1 − δ)(1 + σ) implies ρ* = 0 by Lemma 4, and thus $\phi _{ - 1}^* = 1$ by Lemma 5.

For C’s approval, note χ*⁻ = 1 in all cases, so Lemma 2 implies x _C = 1. ■

Lemma 8. Suppose Δ₂ = 1.

1. 1. If $\gamma \,\le\, \lambda \left( {1 - \delta } \right) \equiv \hat{\gamma }\left( {{\Delta _2} = 1, \cdot } \right)$, then in the most informative PBE, $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right)} \right)$ and ${\rho ^*} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$. C reverses a law approved by G if review is informative and Δ₁ = −1; C upholds a law approved by G otherwise.

2. 2. If $\gamma > \lambda \left( {1 - \delta } \right) \equiv \hat{\gamma }\left( {{\Delta _2} = 1, \cdot } \right)$, then in the most informative PBE, $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {1,1} \right)$ and ρ* = 0. C approves all laws approved by G.

Proof. First, if Δ₂ = 1, γ ≤ λ(1 − δ), and ${\phi _{ - 1}} = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right)$, then by equation (5), C is indifferent between review and no review, so ${\rho ^*} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$ is a best response. If C reviews with probability $\rho = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$, then by equation (9), G is indifferent between upholding and vetoing when Δ₁ = −1, so $\phi _{ - 1}^* = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right)$ is a best response.

Second, if γ > λ(1 − δ), then ρ* = 0 for ϕ ₋₁ = 1 by Lemma 4, so $\phi _{ - 1}^* = 1$ by Lemma 5.

For C’s approval, Lemma 2 implies C will reverse x _G = 1 if Δ₁ = −1 is discovered in review; otherwise, C cannot reverse in this history. ■

Remark. Two facts imply that these are the most informative PBEs in each case. First, from G’s best response [equation (9)], for given sanctions κ _C > 1 and uphold probability χ, G’s utility from approval when Δ₁ = −1 is decreasing (and thus informativeness increasing) in C’s review probability ρ. Given assembly pressure from below, G will not reveal information absent a review threat. However, second, in each parameter region of Lemmas 6–8, C cannot commit to review with probability greater than ρ* specified in each case. Indeed, the primary benefit of increasing salience of legal consistency σ is that σ increases C’s incentive to review—until σ is so great that C does not review at all.

With these results, Proposition 1 from the text is straightforward.

Proposition 1. For all Δ₁, Δ₂, and σ:

1. 1. If $\gamma \,\le\, \hat{\gamma }\left( {{\Delta _2},\sigma } \right)$, then there is a PBE such that $\phi _{ - 1}^* \lt \,\phi _1^*$ and ρ* > 0.

2. 2. If $\gamma > \hat{\gamma }\left( {{\Delta _2},\sigma } \right)$, then in any PBE, $\phi _{ - 1}^* = \phi _1^* = 1$ and ρ* = 0.

Proof. Part 1 is immediate from part 1 of Lemmas 6–8. Part 2 is immediate from part 2 of these lemmas. ■

Proposition 2. For a given review cost γ,

1. 1. If C is sufficiently flexible $\left( {\sigma \,\le\, 2 - {1 \over \delta }} \right)$, then legal inconsistency raises the substantive quality of legislation upheld by G(${\mathbb{E}}\left( {{\Delta _1}{\rm{|}} - 1} \right) \,\ge\, {\mathbb{E}}\left( {{\Delta _1}{\rm{|}}1} \right)\forall \gamma$ and ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}} - 1} \right) \,\ge\, {\mathbb{E}}\left( {{\Delta _1}{\rm{|}}1} \right)$ for some γ).

2. 2. If C is sufficiently fastidious $\left( {\sigma > 2 - {1 \over \delta }} \right)$, then legal inconsistency can reduce the substantive quality of legislation upheld by G (${\mathbb{E}}\left( {{\Delta _1}{\rm{|}} - 1} \right) \lt {\mathbb{E}}\left( {{\Delta _1}{\rm{|}}1} \right)$ for some γ).

Proof. Recall that $\phi _{ - 1}^*\left( {{\Delta _2} = 1} \right) = \min \left\{ {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right)} \right\}$ by Lemma 8. Note that Lemma 1 implies the expected quality of upheld legislation is ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{\Delta _2}} \right) = 2d - 1 = {{\delta - \left( {1 - \delta } \right)\phi _{ - 1}^*} \over {\delta + \left( {1 - \delta } \right)\phi _{ - 1}^*}}$, so it is sufficient to show the effect of Δ₂ on $\phi _{ - 1}^*$.

1. 1. If σ > 2δ − 1 and $\gamma \,\le\, \hat{\gamma }\left( {{\Delta _2} = - 1,\sigma > 2\delta - 1} \right)$, Lemma 6 gives $\phi _{ - 1}^*\left( { - 1} \right) = \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right) \lt \phi _{ - 1}^*\left( {{\Delta _2} = 1} \right)$, so ${\mathbb{E}}\left( {{{\rm{\Delta }}_1}{\rm{|}}{{\rm{\Delta }}_2} = - 1} \right) > {\mathbb{E}}\left( {{{\rm{\Delta }}_1}{\rm{|}}{{\rm{\Delta }}_2} = 1} \right)$.

1. 2. If $\sigma > 2 - {1 / \delta }$, then $\hat{\gamma }\left( {{\Delta _2} = - 1,\sigma > 2\delta - 1} \right) \lt \hat{\gamma }\left( {{\Delta _2} = 1, \cdot } \right)$. If $\hat{\gamma }\left( {{\Delta _2} = - 1,\sigma > 2\delta - 1} \right) \lt \gamma \,\le\, \hat{\gamma }\left( {{\Delta _2} = 1, \cdot } \right)$, Lemmas 6 and 8 give $\phi _{ - 1}^*\left( { - 1} \right) = 1 > \left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda - \gamma }}} \right) = \phi _{ - 1}^*\left( 1 \right)$, so ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{\Delta _2} = - 1} \right) \lt {\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{\Delta _2} = 1} \right)$.

Proposition 3. For a given review cost γ and Δ₂ = −1, the substantive quality of legislation approved by $G,{\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{x_G} = 1} \right)$, increases in legal salience up to $\sigma = \max \left\{ {2\delta - 1,1 - {\gamma \over {\lambda \delta }}} \right\}$.

Proof. Recall that, given Lemma 1, the expected quality of upheld legislation in equilibrium is ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{x_G} = 1} \right) = {{\delta - \left( {1 - \delta } \right)\phi _{ - 1}^*} \over {\delta + \left( {1 - \delta } \right)\phi _{ - 1}^*}}$. Also, note $\sigma = 1 - {\gamma \over {\lambda \delta }} \Leftrightarrow \gamma = \lambda \delta \left( {1 - \sigma } \right) \equiv \hat{\gamma }\left( {{\Delta _2} = - 1,\sigma > 2\delta - 1} \right)$ by Lemma 6.

• • If $2\delta - 1 \lt \,\sigma \,\le\, 1 - {\gamma \over {\lambda \delta }}$, then ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{x_G} = 1} \right) = {{\left( {1 - {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \over {\left( {1 + {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)}}$ by Lemma 6 (part 1). This is increasing in σ.

• • If σ > 2δ − 1 and $\sigma > 1 - {\gamma \over {\lambda \delta }}$, then ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{x_G} = 1} \right) = 2\delta - 1$ by Lemma 6 (part 2). This is constant in σ. Note also $2\delta - 1 \lt {{\left( {1 - {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \over {\left( {1 + {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)}}$.

• • If ${{\gamma - \lambda \left( {1 - \delta } \right)} \over {\lambda \left( {1 - \delta } \right)}} \,\le\, \sigma \,\le\, 2\delta - 1$, then ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{x_G} = 1} \right) = {{\left( {1 - {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \over {\left( {1 + {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)}}$ by Lemma 7 (part 1). This is increasing in σ.

• • If σ ≤ 2δ − 1 but $\sigma > {{\gamma - \lambda \left( {1 - \delta } \right)} \over {\lambda \left( {1 - \delta } \right)}}$, then ${\mathbb{E}}\left( {{\Delta _1}{\rm{|}}{x_G} = 1} \right) = 2\delta - 1$ by Lemma 7 (part 2). This is constant in σ. Note also $2\delta - 1 \lt {{\left( {1 - {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \over {\left( {1 + {\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)}}$. ■

APPENDIX B: SANCTIONS ON G FOR ANY REVERSAL

This appendix analyzes the model in which κ _C is applied whenever x _G = 1 but x _C = 0, i.e., C sanctions for any reversal. The rest of the model is unchanged. Call this model S _A. The model of the main text, with sanctions only after Crown review and reversal, is S _R.

First consider the effect of S _A on best responses.

• • S _A does not change C’s utility from approval, conditional on Δ₂, C’s posterior d, and the informativeness of review Λ. In particular, Lemma 2 shows that for S _A, if C discovers Δ₁ in review, x _C = 1 if and only if Δ₁ = 1.

• • S _A does not change C’s utility of review conditional on Δ₂, C’s posterior d, and G’s action x _G.

Together with C’s acceptance behavior noted above, C’s review strategy generates an ex ante probability of x _C = 1. Denote these probabilities by ${\chi _{{\Delta _1}}} = \left( {{\chi _{ - 1}},{\chi _1}} \right)$. These acceptance probabilities are a function of Δ₁ because C can discover Δ₁ in review, and x _C is based on that discovery.

• • G’s approval if Δ₁ = −1: x _G = 1 confers ${\mathbb{E}}\;U_G^1 = - {\chi _{ - 1}} - \left( {1 - {\chi _{ - 1}}} \right){\kappa _C} = {\chi _{ - 1}}\left( {{\kappa _C} - 1} \right) - {\kappa _C}$, since crown sanctions are applied whenever x _G is overturned. ${\mathbb{E}}\;U_G^0 = - {\kappa _A}$. So G will mix if ${\chi _{ - 1}} = {\hat{\chi }_{ - 1}} \equiv \left( {{{{\kappa _C} - {\kappa _A}} \over {{\kappa _C} - 1}}} \right)$. G will approve for ${\chi _{ - 1}} > {\hat{\chi }_{ - 1}}$ and veto for ${\chi _{ - 1}} \lt {\hat{\chi }_{ - 1}}$.

• • G’s approval if Δ₁ = 1: x _G = 1 confers ${\mathbb{E}}\;U_G^1 = {\chi _1} - \left( {1 - {\chi _1}} \right){\kappa _C} = {\chi _1}\left( {{\kappa _C} + 1} \right) - {\kappa _C}$. ${\mathbb{E}}\;U_G^0 = - {\kappa _A}$. So G will mix if ${\chi _1} = {\hat{\chi }_1} \equiv \left( {{{{\kappa _C} - {\kappa _A}} \over {{\kappa _C} + 1}}} \right)$. G will approve for ${\chi _1} > {\hat{\chi }_1}$ and veto for ${\chi _1} \lt {\hat{\chi }_1}$.

PBE places restrictions on the absolute and relative magnitudes of (χ ₋₁, χ ₁).

Lemma 9. In any PBE under S _A, χ ₁ ≥ χ ₋₁ (crown monotonicity).

Proof. Let χ ⁺(Δ₁) denote the probability that C approves when informed of the state and the state is Δ₁. Let χ ⁻ denote the probability that C approves when uninformed of the state. By the law of total probability, ${\chi _{{\Delta _1}}} = \rho \lambda {\chi ^ + }\left( {{\Delta _1}} \right) + \left( {1 - \rho \lambda } \right){\chi ^ - }$. Since χ ⁺(1) = 1, χ ⁺(−1) = 0, and χ ⁻ is not a function of Δ₁, it follows that χ ₁ ≥ χ ₋₁. ■

Note that for Δ₂ = 1, χ ⁻ = 1 by construction. For Δ₂ = −1, χ ⁻ < 1 is possible.

Lemma 10. In any PBE under S _A, χ ₁ ∈ {0, ρλ, 1}, and χ ₋₁ ∈ {0, 1 − ρλ, 1}.

Proof. From Lemma 9, χ ₁ = ρλ + (1 − ρλ)χ ⁻. So ρ > 0 ⇒ χ ₁ ≥ ρλ. If Δ₂ = −1 and ${\mathbb{E}}\left[ {{\Delta _1}{\rm{|}}{x_G} = 1} \right] \,\lt\, \sigma$ then χ ⁻ = 0, so χ ₁ = ρλ. If χ ⁻ = ρ = 0, then χ ₁ = 0. If either Δ₂ = −1 and ${\mathbb{E}}\left[ {{\Delta _1}{\rm{|}}{x_G} = 1} \right] \,\ge\, \sigma$, or Δ₂ = 1, then χ ⁻ = 1, so χ ₁ = ρλ + (1 − ρλ) = 1.

Further, χ ₋₁ = (1 − ρλ)χ ⁻. If Δ₂ = −1 and ${\mathbb{E}}\left[ {{\Delta _1}{\rm{|}}{x_G} = 1} \right] \lt \sigma$ then χ ⁻ = 0, so χ ₋₁ = 0. If Δ₂ = −1 and ${\mathbb{E}}\left[ {{\Delta _1}{\rm{|}}{x_G} = 1} \right] \,\ge\, \sigma$, or if Δ₂ = 1, then χ ⁻ = 1, so χ ₋₁ = 1 − ρλ. If, in addition, ρ = 0, then χ ₋₁ = 1.

Note that χ ⁻ = 0 is not possible when Δ₂ = 1 by construction.

An immediate implication is that there is no PBE with χ ₋₁ = 1 − ρλ and χ ₁ = ρλ. The latter is possible only if χ ⁻ = 1, but that implies χ ₁ = 1.

Lemma 11. There is no PBE with χ ₁ = 1 and χ ₋₁ = 0.

Proof. Suppose χ ₁ = 1 and χ ₋₁ = 0. Then ${\chi _1} > {\hat{\chi }_1}$ and ${\chi _{ - 1}} \lt {\hat{\chi }_{ - 1}}$, so $\phi _1^* = 1$ and $\phi _{ - 1}^* = 0$. But then ρ* = 0 if γ > 0. Then χ ₁ = 1 ⇒ χ ⁻ = 1 by Lemma 10. But ρ = 0 and χ ⁻ = 1 imply χ ₋₁ = 1, a contradiction. ■

Thus, there is no PBE with full separation by Δ₁. An immediate implication is that there is no PBE with ρλ = 1, because this implies χ ₁ = 1 and χ ₋₁ = 0.

Lemma 12. In any PBE under S _A, if χ ₋₁ = 0, then $\phi _1^* = \phi _{ - 1}^* = 0$.

Proof. Note that ${\chi _{ - 1}} = 0 \lt {\hat{\chi }_{ - 1}}$, so ${\chi _{ - 1}} = 0 \,\Rightarrow\, \phi _{ - 1}^* = 0$.

For χ ₁, suppose first that χ ₁ = 0. Then, ${\chi _1} = 0 \,\lt\, {\hat{\chi }_1}$, so $\phi _1^* = 0$.

Suppose second that χ ₁ > 0. In addition, from Lemmas 10 and 11, χ ₋₁ = 0 implies χ ⁻ = 0. But then χ ⁻ = 0 ⇒ χ ₁ = ρλ, from which χ ₁ > 0 ⇒ ρ > 0.

If $\rho \,\ge {{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} + 1} \right)}}$, then ${\chi _1} \ge \,{\hat{\chi }_1}$, so $\phi _1^* = 1$. But if ϕ ₋₁ = 0 and ϕ ₁ > 0, then ${\mathbb{E}}\left[ {{\Delta _1}{\rm{|}}{x_G} = 1} \right] \,\ge\, \sigma$, so χ ⁻ = 1; thus χ ₋₁ > 0, a contradiction. If $\rho \,\lt {{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} + 1} \right)}}$, then ${\chi _1} \lt {\hat{\chi }_1}$, so $\phi _1^* = 0$. ■

Thus, either C must risk approving when Δ₁ = −1, or G never approves.

Given these results, for Δ₂ = −1, the only PBE-feasible (χ ₋₁, χ ₁) pairs are:

1. 1. $\left( {0,\rho \lambda \lt {{{\kappa _A} - 1} \over {{\kappa _C} + 1}}} \right)$: G preemptively vetoes everything.

2. 2. (1, 1): C accepts everything without review.

3. 3. (1 − ρλ, 1): C accepts unless hard evidence of Δ₁ = −1, then rejects.

For Δ₂ = 1, only (χ ₋₁, χ ₁) = (1, 1) or (1 − ρλ, 1) are possible. All other (χ ₋₁, χ ₁) pairs violate either Lemma 9, 10, or 11.

Now consider PBE under the incentive feasible (χ ₋₁, χ ₁) pairs in turn. Lemma 6 requires significant modification in both parts, but Lemmas 7 and 8 hold without modification under S _A.

Lemma 6′. Suppose Δ₂ = −1 and σ > 2δ − 1.

1. 1. If $\gamma \,\le\, {{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2}$, then in the most informative PBE under S _A, $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \right)$ and ${\rho ^{\rm{*}}} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$. C reverses a law approved by G if review is informative and Δ₁ = −1; C upholds a law approved by G otherwise.

2. 2. If $\gamma > {{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2}$, then in the most informative PBE, $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {0,0} \right)$ and ρ* = 0. C believes Pr[Δ₁ = 1|x _G = 1] = δ and would reverse any law approved by G.

Proof. Part 1. Note that $\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right) = \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)$ for $\gamma = {{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2} \lt\, \lambda \left( {1 - \delta } \right)\left( {1 + \sigma } \right)$. For any smaller γ, C is indifferent about review for $\left( {{\phi _1},{\phi _{ - 1}}} \right) = \left( {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \right)$. Therefore, ${\rho ^*} = \left( {{{{\kappa _A} - 1} \over {\lambda \left( {{\kappa _C} - 1} \right)}}} \right)$ and χ ⁻ = 1 is a best response for C. This implies ${\chi _1} = 1 > {\hat{\chi }_1}$ and ${\chi _{ - 1}} = 1 - {\rho ^*}\lambda = {\hat{\chi }_{ - 1}}$. Note also that any ${\phi _{ - 1}} > \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)$ cannot be optimal since it induces χ ₋₁ = 0 and sanction κ _C with certainty. So $\left( {\phi _1^*,\phi _{ - 1}^*} \right) = \left( {1,\left( {{\delta \over {1 - \delta }}} \right)\left( {{\gamma \over {\lambda \left( {1 + \sigma } \right) - \gamma }}} \right)} \right)$ is a best response for G.

Part 2. Pr[Δ₁ = 1|x _G = 1] = δ implies ${\mathbb{E}}\left[ {{\Delta _1}{\rm{|}}{x_G} = 1} \right] = 2\delta - 1$. Given σ > 2δ − 1, equation (4) implies χ ⁻ = 0 is a best response. Further, for Pr[Δ₁ = 1|x _G = 1] = δ, equation (7) implies that C reviews only if γ < λδ(1 − σ). But ${{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2} > \lambda \delta \left( {1 - \sigma } \right)$, so ρ* = 0. Since ρ* = 0 and χ ⁻ = 0, it follows that χ ₋₁ = 0 by Lemma 10, so $\phi _{ - 1}^* = \phi _1^* = 0$ by Lemma 12. Then x _G = 1 is off the equilibrium path, and Pr[Δ₁ = 1|x _G = 1] = δ is Bayes consistent.^{Footnote 37} ■

Thus, Part (1) of Lemma 6 changes under S _A. For λδ(1 − σ) < γ ≤ λ(1 − δ)(1 + σ), $\phi _{ - 1}^* = 1$ is no longer optimal for G. This exposes G to the sanction −κ _C with certainty. G prefers to mix when Δ₁ = −1, incurring −κ _C and −κ _A probabilistically, until γ is so large that C will reject without review even when G mixes. Part (2) of Lemma 6 also does not hold under S _A: instead of G approving all laws in this case, G vetoes all laws preemptively to avoid crown sanctions.

Under part 1, S _A generates more information about Δ₁ than S _R. Under part 2, S _A is inefficient relative to S _R. It imposes additional costs on G from assembly sanctions without changing policy outcomes. The reason for the inefficiency is that G cares not only about the disposition of the law and costs of reversal, but about which actor makes a decision to reject. G incurs pressure from A if G itself rejects, but not if C rejects.

Lemma 7′. Suppose Δ₂ = −1 and σ ≤ 2δ − 1. Lemma 7 holds under S _A.

Proof. Part 1: If (χ ₋₁, χ ₁) = (1 − ρλ, 1), then $\phi _1^* = 1$. Given this, the posterior d is as in equation (1), and C reviews only if ${\phi _{ - 1}} \le \left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right)$ and γ ≤ λ(1 − δ)(1 + σ). If σ ≤ 2δ − 1, then $\left( {{\delta \over {1 - \delta }}} \right)\left( {{{1 - \sigma } \over {1 + \sigma }}} \right) > 1$, and best responses are as in part 1 of Lemma 7.

Part 2. If (χ ₋₁, χ ₁) = (1, 1) (summary approval), G pools on $\phi _{ - 1}^* = \phi _1^* = 1$. If σ ≤ 2δ − 1 and γ > λ(1 − δ)(1 + σ), χ ₋₁ = χ ₁ = 1 is a best response. Thus, part (2) of Lemma 7 holds under both S _A and S _R. ■

Lemma 8′. Suppose Δ₂ = 1. Lemma 8 holds under S _A.

Proof. The original proof holds without modification. In particular, χ ₁ = 1 under S _A, which implies $\phi _1^* = 1$, and S _A does not change C’s best response conditional on Δ₂ = 1 and $\phi _1^* = 1$. ■

In summary,

• • S _A and S _R generate the same PBE behavior for:

  • – Δ₂ = 1

  • – Δ₂ = −1 and σ ≤ 2δ − 1

  • – Δ₂ = −1, σ > 2δ − 1, and γ < λδ(1 − σ)

  • – Δ₂ = −1, σ > 2δ − 1, and γ > λ(1 − δ)(1 + σ)

• • S _A elicits more information about Δ₁ than S _R for

  • – Δ₂ = −1, σ > 2δ − 1, and $\gamma \in \left( {\lambda \delta \left( {1 - \sigma } \right),{{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2}} \right] \equiv {\Gamma _ - }$

• • S _A is inefficient relative to S _R for

  • – Δ₂ = −1, σ > 2δ − 1, and $\gamma \in \left( {{{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2},\lambda \left( {1 - \delta } \right)\left( {1 + \sigma } \right)} \right] \equiv {\Gamma _ + }$

These regions have width $\left| {{\Gamma _ + }} \right| = {{\lambda \left( {1 + \sigma } \right)\left( {\sigma - 2\delta + 1} \right)} \over 2}$ and $\left| {{\Gamma _ - }} \right| = {{\lambda \left( {1 - \sigma } \right)\left( {\sigma - 2\delta + 1} \right)} \over 2}$. So the inefficiency region is never smaller. For σ > 2δ − 1, the relative width is ${{1 + \sigma } \over {1 - \sigma }} > 1$, which is increasing in σ. For σ = 2δ − 1, |Γ₊| = |Γ₋| = 0. For σ → 1, |Γ₋| → 0 and |Γ₊| → 2λ(1 − δ).

For the following result, suppose that C internalizes cost ψκ _A when G vetoes for γ ∈ Γ₊—the case where G’s veto has no effect on policy. This captures, in a reduced form way, that these costs affect the participation constraint of potential governors in a larger model with endogenous recruitment, and this is costly for C. Assume also that γ is uniformly distributed over [λδ(1 − σ), λ(1 − δ)(1 + σ)], the region over which S _A changes behavior. Let V(S) denote the expected utility under cost regime S.

Proof. V(S _R) = V(S _A) under Δ₂ = 1, and under Δ₂ = −1 and σ ≤ 2δ − 1, by Lemmas 7 and 8.

Now assume Δ₂ = −1 and σ ∈ (2δ − 1, 1).

For S _R, V(S _R) = 0, because C summary reverses in this range.

For S _A, let V(Γ_i, S _A) denote V’s expected utility under regime S _A for γ ∈ Γ_i, i ∈ {+, −}. Then V(Γ₊, S _A) = −ψκ _A. In this region, G preemptively vetoes under S _A to avoid summary veto by C, but then incurs assembly sanctions, a portion of which C internalizes. On the other hand,

$$V\left( {{\Gamma _ - },{S_A}} \right) = \delta \left( {1 - \sigma } \right) - \mathop \int \nolimits _{\lambda \delta \left( {1 - \sigma } \right)}^{{{\lambda \left( {1 + \sigma } \right)\left( {1 - \sigma } \right)} \over 2}}\left( {\left( {\left( {1 - \delta } \right)\left( {1 - {\rho ^*}\lambda } \right)\phi _{ - 1}^*\left( \gamma \right)\left( {1 + \sigma } \right)} \right) + \left( {{\rho ^*}\gamma } \right)} \right){\rm{d}}F\left( \gamma \right).$$

The terms are as follows. First, if Δ₁ = 1, then $\phi _1^* = 1$ and C always upholds, obtaining utility (1 − σ). This occurs with probability δ. Second, if Δ₁ = −1, and G upholds, and C does not review or it is not informative, C obtains utility (−1 − σ). This occurs with probability $\left( {1 - \delta } \right)\left( {1 - {\rho ^*}\lambda } \right)\phi _{ - 1}^*$. Third, C incurs cost γ for review, which occurs with probability ρ*.

Combining these, V(S _A) = Pr[γ ∈ Γ₋]V(Γ₋, S _A) + Pr[γ ∈ Γ₊]V(Γ₊, S _A) or $\left( {{{1 - \sigma } \over 2}} \right)V\left( {{\Gamma _ - },{S_A}} \right) + \left( {{{1 + \sigma } \over 2}} \right)V\left( {{\Gamma _ + },{S_A}} \right)$.

If ψ = 0, then $V\left( {{S_A}} \right) = \left( {{{1 - \sigma } \over 2}} \right)V\left( {{\Gamma _ - },{S_A}} \right) > V\left( {{S_R}} \right) = 0$ for all 2δ − 1 < σ < 1 (otherwise C would not choose to review over Γ₋). However, V(Γ₋, S _A) < δ(1 − σ) and Pr[γ ∈ Γ₋] < Pr[γ ∈ Γ₊] ∀σ ∈ (2δ − 1, 1). Therefore, V(S _A) < 0 = V(S _R) for $\psi \,\ge\, {{\delta \left( {1 - \sigma } \right)} \over {{\kappa _C}}} \equiv \,\hat{\psi }\left( \sigma \right)$. Since σ > 2δ − 1, it follows that $\hat{\psi } \,\lt {{2\delta \left( {1 - \delta } \right)} \over {{\kappa _A}}} \lt {1 \over 2}$, ${{{\rm{d}}\hat{\psi }} \over {{\rm{d}}\sigma }} \lt 0$, and ${\lim _{\sigma \to 1}}\hat{\psi } = 0$.