Proof. Let $\mathbb {A} = \langle \textbf {A}, \wedge , \vee , \Rightarrow , \textbf {1}, \textbf {0} \rangle $ be a $\mathsf {WoDRIA}$ . Then, $\mathbb {A}$ is a complete distributive lattice. We shall prove that all the properties P1, P2, P3, and P4 will be satisfied by $\mathbb {A}$ .

For P1: Let $a, b, c \in \textbf {A}$ be such that $a \wedge b \leq c$ . We have to prove that $a \leq b \Rightarrow c$ . If $b \Rightarrow c = \textbf {1}$ , then we are done. Let $b \Rightarrow c = \textbf {0}$ , which implies that $b \neq \textbf {0}$ and $c = 0$ . Therefore, by our assumption, $a = \textbf {0}$ . Hence, in any case, we conclude that $a \leq b \Rightarrow c$ .

For P2: Let $a \leq b$ holds in $\textbf {A}$ . We have to prove that $c \Rightarrow a \leq c \Rightarrow b$ , for any $c \in \textbf {A}$ . If $c \Rightarrow b = \textbf {1}$ , the proof is done. Suppose $c \Rightarrow b = \textbf {0}$ . Then, by the definition of $\Rightarrow $ , $c \neq \textbf {0}$ and $b = \textbf {0}$ . Since $a \leq b$ , we have $a = \textbf {0}$ also. Hence, it can be concluded that $c \Rightarrow a = \textbf {0}$ . Therefore, $c \Rightarrow a \leq c \Rightarrow b$ , for any $c \in \textbf {A}$ .

For P3: Let us consider any $a, b \in \textbf {A}$ which satisfy the property $a \leq b$ . We shall prove that for any $c \in \textbf {A}$ , $b \Rightarrow c \leq a \Rightarrow c$ holds. Similar to the previous cases, there is nothing to prove if $a \Rightarrow c = \textbf {1}$ . Hence, assume that $a \Rightarrow c = \textbf {0}$ . This implies that $a \neq \textbf {0}$ and $c = \textbf {0}$ . Since, by our assumption, $a \leq b$ , we get that $b \neq \textbf {0}$ as well. Hence, $b \Rightarrow c = \textbf {0}$ . So, in any case, $b \Rightarrow c \leq a \Rightarrow c$ holds for any $c \in \textbf {A}$ .

For P4: Let us consider any three elements $a, b, c \in \textbf {A}$ . If $(a \wedge b) \Rightarrow c = \textbf {0}$ , then $a, b \neq \textbf {0}$ and $c = \textbf {0}$ , which entails that $a \Rightarrow (b \Rightarrow c) = \textbf {0}$ . Conversely, if $a \Rightarrow (b \Rightarrow c) = \textbf {0}$ , we have $a, b \neq \textbf {0}$ and $c = \textbf {0}$ . Since $\leq $ is a well-order (and hence linear-order) relation in $\textbf {A}$ , either $a \wedge b = a$ or $a \wedge b = b$ . In either case, $a \wedge b \neq \textbf {0}$ . Therefore, $(a \wedge b) \Rightarrow c = \textbf {0}$ . Hence, we can conclude that, $(a \wedge b) \Rightarrow c = \textbf {0}$ iff $a \Rightarrow (b \Rightarrow c) = \textbf {0}$ . Since the range of the binary operator $\Rightarrow $ is defined to be $\{\textbf {1}, \textbf {0}\}$ , from the previous conclusion, we get that $(a \wedge b) \Rightarrow c = \textbf {1}$ iff $a \Rightarrow (b \Rightarrow c) = \textbf {1}$ . Combining both the results we finally have $(a \wedge b) \Rightarrow c = a \Rightarrow (b \Rightarrow c)$ .⊣

Observation 7. For any $\mathcal {PS}$ -algebra $\mathbb {A}$ and any designated set D, $\textbf{V}^{(\mathbb {A})} \models \mathrm {NFF}$ - $\mathrm {ZF}$ , by Theorem 5, since any $\mathcal {PS}$ -algebra is by definition a $\mathsf {WoDRIA}$ . Moreover, if the logic $\textbf {L}(\mathbb {A}, D)$ is non-classical then $\textbf{V}^{(\mathbb {A})}$ becomes an algebra-valued model of a non-classical set theory.

Proof. That the relation $\sim $ is reflexive and symmetric follows immediately from the definition of . We shall now prove that $\sim $ is transitive as well. Let us consider three elements $u, v, w \in \textbf{V}^{(\mathbb {A})}$ such that $u \sim v$ and $v \sim w$ hold, i.e., . Hence, we get that , as the only possible values of and are $\textbf {1}$ and $\textbf {0}$ . Then, it is enough to prove that as well. If not, then without loss of generality, let there exists $x_0 \in \mathrm {dom}(u)$ such that . Hence, $u(x_0) \neq \textbf {0}$ and . Since, by our assumption, and $u(x_0) \neq \textbf {0}$ , we have . Therefore, there exists $y_0 \in \mathrm {dom}(v)$ such that $v(y_0) \neq \textbf {0}$ and . Hence, we can derive that , as . This implies that . Hence, we get that , which contradicts our assumption. Finally we can conclude that . Hence, $u \sim w$ holds. ⊣

Proof. Following the proof of Theorem 13 of [Reference Tarafder, Banerjee and Krishna11], one can prove that if $u \in \textbf{V}^{(\mathbb {A})}$ is an $\alpha $ -like element for some $\alpha \in \mathrm {ORD}$ , then $\textbf{V}^{(\mathbb {A})} \models \mathsf {Ord}(u)$ .

Conversely, let $\textbf{V}^{(\mathbb {A})} \models \mathsf {Ord}(u)$ for some $u \in \textbf{V}^{(\mathbb {A})}$ . We shall prove that u is an $\alpha $ -like element for some $\alpha \in \mathrm {ORD}$ . If $\mathrm {dom}_D(u)$ is empty then u is a 0-like element and the proof is complete. Hence, we assume that $\mathrm {dom}_D(u)$ is non-empty. Since $\textbf{V}^{(\mathbb {A})} \models \mathrm {Trans}(u)$ , if $v \in \mathrm {dom}_D(u)$ and $w \in \mathrm {dom}_D(v)$ then $w \in \mathrm {dom}_D(u)$ . This leads to the fact that there exists a 0-like element in $\mathrm {dom}_D(u)$ . Also, if there exists a $\beta $ -like element in $\mathrm {dom}_D(u)$ for some $\beta \in \mathrm {ORD}$ , then for every $\gamma \in \beta $ there exists a $\gamma $ -like element in $\mathrm {dom}_D(u)$ . Using this fact and the case that every element of $\textbf{V}^{(\mathbb {A})}$ is sets (not proper classes) in $\textbf{V}$ , we can say that there exists $\alpha \in \mathrm {ORD}$ such that for every $\beta \in \alpha $ , there exists at least one $\beta $ -like element in $\mathrm {dom}_D(u)$ , but there does not exist any $\gamma $ -like element in $\mathrm {dom}_D(u)$ where $\gamma = \alpha $ or $\alpha \in \gamma $ . If $\mathrm {dom}_D(u)$ does not contain any other element then u is an $\alpha $ -like element and hence the proof is complete. If possible, let $\mathrm {dom}_D(u)$ contain an element $x_i$ which is not a $\beta $ -like element for any $\beta \in \alpha $ . Since $\textbf{V}^{(\mathbb {A})} \models \mathrm {LO}(u)$ , $\mathrm {dom}_D(x_i)$ contains a $\beta $ -like element for each $\beta \in \alpha $ . But, by a similar argument, it contains an element $x_j$ which is not $\beta $ -like for any $\beta \in \alpha $ . But $x_j \in \mathrm {dom}_D(u)$ . Continuing the process, we get a subset $\{ x_i: i \in \Lambda \}$ of $\mathrm {dom}_D(u)$ , where $\Lambda $ is an index set such that

1. (i) for every $i \in \Lambda $ , $x_i$ is not $\beta $ -like for any $\beta \in \alpha $ , and

2. (ii) for every $i \in \Lambda $ , $x_i$ contains a $\beta $ -like element for each $\beta \in \alpha $ .

Now consider an element $y \in \textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}(y) = \{ x_i: i \in \Lambda \}$ and $\mathrm {ran}(y) = \{\textbf {1}\}$ . Then $\textbf{V}^{(\mathbb {A})} \models y \subseteq u ~ \wedge ~ (y \neq \varnothing )$ . Since $\textbf{V}^{(\mathbb {A})} \models \mathrm {WO}_\in (u)$ , the second conjunct of $\mathrm {WO}_\in (u)$ is also valid in $\textbf{V}^{(\mathbb {A})}$ . So, there exists an element $z \in \textbf{V}^{(\mathbb {A})}$ such that for some $k \in \Lambda $ , and $\mathrm {dom}_D(z)$ does not contain any element which is equal to $x_i$ for any $i \in \Lambda $ . This fact and the construction of $x_k$ show that $\mathrm {dom}_D(x_k)$ contains at least one $\beta $ -like element for each $\beta \in \alpha $ and nothing else. Hence, from the definition, we get that $x_k$ is an $\alpha $ -like element. Moreover, by our construction, $x_k \in \mathrm {dom}_D(u)$ . This contradicts the fact that $\mathrm {dom}_D(u)$ does not contain any $\alpha $ -like element. Hence, the proof is complete. ⊣

Proof. The theorem will be proved by meta-induction. Let us arbitrarily fix a $\mathcal {PS}$ -algebra $\mathbb {A}$ with a designated set D.

Base step: Consider the case for $n = 0$ . We need to prove that the 1-like elements are the only instances of the formula

$$ \begin{align*}\mathsf{Nat}(1,x): = \exists x_0 \; \big{(}\mathsf{Nat}(0,x_0) \; \wedge \; x_0 \in x \; \wedge \; \forall y \; (y \in x \rightarrow y = x_0)\big{)}.\end{align*} $$

So $\textbf{V}^{(\mathbb {A})} \models \mathsf {Nat}(1,x)$ if and only if there exists $u \in \textbf{V}^{(\mathbb {A})}$ such that , , .

1. (i) if and only if u is 0-like.

2. (ii) , i.e., , if and only if there exists $t \in \mathrm {dom}_D(x)$ such that , i.e., t is 0-like, using Theorem 9(i).

3. (iii) , i.e., , if and only if for each $t \in \mathrm {dom}_D(x)$ , , i.e., t is 0-like.

Hence, combining (i), (ii), and (iii), it can be concluded that $\textbf{V}^{(\mathbb {A})} \models \mathsf {Nat}(1, u)$ if and only if u is a $1$ -like element.

Induction hypothesis: Let the proposition be true for all natural numbers less than $m \in \omega \setminus \{0\}$ , i.e., for $u \in \textbf{V}^{(\mathbb {A})}$ , $\textbf{V}^{(\mathbb {A})} \models \mathsf {Nat}(k, u)$ if and only if u is a k-like element for each $k \in \{1, 2, \ldots , m\}$ .

Induction step: We shall prove the proposition for the natural number m. Hence, we have to prove $\textbf{V}^{(\mathbb {A})} \models \mathsf {Nat}(m+1, u)$ if and only if u is an ( $m+1$ )-like element. Since

$$ \begin{align*} \mathsf{Nat}(m+1,u) :=~ & \exists x_0 \; \exists x_1 \ldots \exists x_m \big{(}(\mathsf{Nat}(0,x_0) \wedge \; \cdots \; \wedge \mathsf{Nat}(m,x_m)) \wedge\\& (x_0\kern-1pt \in\kern-1pt u \kern-1.5pt\; \wedge \kern-1.5pt\; \cdots \kern-1.5pt\; \wedge \kern-1.5pt\; x_m \kern-1pt\in\kern-1pt u) \kern-1.5pt\; \wedge \kern-1.5pt\; \forall y (y \kern-1pt\in\kern-1pt u \rightarrow\kern-1pt (y \kern-1pt =\kern-1pt x_0 \kern-1.5pt\; \vee \kern-1.5pt\; \cdots\kern-1.5pt \; \vee \kern-1.5pt\; y \kern-1pt=\kern-1pt x_m)\kern-.8pt)\kern-.8pt\big{)}, \end{align*} $$

if and only if there exist $u_0, u_1, \ldots , u_m \in \textbf{V}^{(\mathbb {A})}$ such that

1. (i) ,

2. (ii) , and

3. (iii) .

By the induction hypothesis, (i) holds if and only if each $u_i$ is an i-like element where $i \in \{0, 1, \ldots , m\}$ . Condition (ii) holds if and only if for each $i \in \{0, 1, 2, \ldots , m\}$ there exists an i-like element, in $\mathrm {dom}_D(u)$ . Condition (iii) holds if and only if every $y \in \mathrm {dom}_D(u)$ is i-like for some $i \in \{0, 1, 2, \ldots , m\}$ . Hence, if and only if u is an ( $m+1$ )-like element, which entails that the proposition also holds for m. Therefore, by the principle of mathematical induction, the proposition holds for all $n \in \omega $ . ⊣

Proof. Let us take an n-like element x for some $n \in \omega $ . Now let

From the first conjunct we get

, which implies that for any $y \in \textbf{V}^{(\mathbb {A})}$ ,

, i.e.,

i.e., any $z \in \mathrm {dom}_D(y)$ is either an m-like element, where $m < n$ , or an n-like element.

The second conjunct gives

This implies that for any $y \in \textbf{V}^{(\mathbb {A})}$ ,

, i.e.,

i.e., for any $z \in \textbf{V}^{(\mathbb {A})}$ which is either m-like for some $m < n$ or n-like, there exists $t \in \mathrm {dom}_D(y)$ which is either m-like or n-like.

Combining the above two derivations, we can say that if x is any n-like element and holds, then by definition, y is an $(n+1)$ -like element. This completes the proof. ⊣

Proof. If possible, let u be an $(m-1)$ -like element such that $u \in \mathrm {dom}_D(I)$ . We have

Condition (i) and $I(u) \in D$ together imply

i.e., there exists $v \in \textbf{V}^{(\mathbb {A})}$ such that

which implies

. Using Theorem 9 the first conjunct assures that v is m-like. The second conjunct implies $I(v) \in D$ . This contradicts condition (ii). Hence, the proof is complete.⊣

Proof. Let us consider an $I \in \textbf{V}^{(\mathbb {A})}$ such that $\textbf{V}^{(\mathbb {A})} \models \mathrm {Ind}(I)$ , i.e., both the conjuncts of $\mathrm {Ind}(I)$ are valid in $\textbf{V}^{(\mathbb {A})}$ . Hence, from the first conjunct there exists a 0-like element in $\mathrm {dom}_D(I)$ . Since the second conjunct is also valid, by Lemma 16 it can be concluded that for a natural number m in $\textbf{V}$ , if there exists an m-like element $u \in \mathrm {dom}_D(I)$ then there exists an $(m+1)$ -like element $v \in \mathrm {dom}_D(I)$ . Hence, by meta-induction on natural numbers in $\textbf{V}$ , the proof is complete.⊣

Proof. Let u be an n-like element for some $n \in \omega $ . We get

since for each $I \in \textbf{V}^{(\mathbb {A})}$ , if

then by Lemma 17 we also get

.

Conversely, let . Hence, for each $I \in \textbf{V}^{(\mathbb {A})}$ , if then . We shall show that x is a natural-number-like element. Consider an $\omega $ -like element $u \in \textbf{V}^{(\mathbb {A})}$ . Then is immediate. But if x is not a natural-number-like element then using Lemma 17 we get . This implies that x is a natural-number-like element and the theorem is proved.⊣

Proof. Consider any $x, y \in \textbf{V}^{(\mathbb {A})}$ such that

. We shall prove

, i.e., y is an $\omega $ -like element. Using Theorem 19,

implies that x is an $\omega $ -like element. From the second conjunct we have

, which implies

Since x is $\omega $ -like, if there exists $t \in \mathrm {dom}(y)$ which is not a natural-number-like element then $y(t) \notin D$ . The third conjunct of our assumption gives us that

. Then, using Lemma 17, we get that for each natural number n there exists an n-like element in $\mathrm {dom}_D(y)$ . These results together show that y is an $\omega $ -like element. ⊣

Proof. Let us consider a $\mathcal {PS}$ -algebra $\mathbb {A}$ with a designated set D and a model of set theory $\textbf{V}$ such that $\textbf{V} \models \mathrm {AC}$ . Suppose $u \in \textbf{V}^{(\mathbb {A})}$ be an arbitrary element such that $\textbf{V}^{(\mathbb {A})} \models u \neq \varnothing $ . Then, we have $\mathrm {dom}_D(u) \neq \varnothing $ in $\textbf{V}$ . Let us take an element $\bar {x} \in \mathrm {Part}(\mathrm {dom}_D(u))$ .

Case I. Suppose $\bar {x}$ does not contain 0-like elements. Arbitrarily fix an element $a_{\bar {x}} \in \bar {x}$ . By our assumption $\mathrm {dom}_D(a_{\bar {x}}) \neq \varnothing $ and hence choose an element $b_{\bar {x}} \in \mathrm {dom}_D(a_{\bar {x}})$ . Construct two elements $p_{\bar {x}}$ and $q_{\bar {x}}$ in $\textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}(p_{\bar {x}}) = \{a_{\bar {x}}\}$ and $\mathrm {dom}(q_{\bar {x}}) = \{a_{\bar {x}}, b_{\bar {x}}\}$ , where $\mathrm {ran}(p_{\bar {x}}) = \{\textbf {1}\} = \mathrm {ran}(q_{\bar {x}})$ . Using these elements $p_{\bar {x}}$ and $q_{\bar {x}}$ we define another element $w_{\bar {x}} \in \textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}(w_{\bar {x}}) = \{p_{\bar {x}}, q_{\bar {x}}\}$ and $\mathrm {ran}(w_{\bar {x}}) = \{\textbf {1}\}$ . Hence by Definition 21, it can be said that $w_{\bar {x}}$ is an $(a_{\bar {x}}, b_{\bar {x}})$ -like element.

Case II. Suppose $\bar {x}$ contains a 0-like element. Then by Theorem 9(i), all the elements of $\bar {x}$ are 0-like elements. Hence, $\bar {x}$ is the class of all 0-like elements of $\mathrm {dom}_D(u)$ . Let us arbitrarily fix any two 0-like elements $s, t \in \textbf{V}^{(\mathbb {A})}$ , not necessarily different in $\textbf{V}$ . Following the same construction as in Case I, construct an element $w_{\bar {x}} \in \textbf{V}^{(\mathbb {A})}$ such that it becomes an $(s, t)$ -like element.

Let us now consider an element f such that $\mathrm {dom}(f) = \{w_{\bar {x}}~:~\bar {x} \in \mathrm {Part}(\mathrm {dom}_D(u))\}$ and $\mathrm {ran}(f) = \{\textbf {1}\}$ . The existence of f in $\textbf{V}$ is assured by the fact that $\textbf{V} \models \mathrm {AC}$ . Then by the construction $f \in \textbf{V}^{(\mathbb {A})}$ . It can be checked that $\textbf{V}^{(\mathbb {A})} \models \mathrm {Func}(f) \wedge \mathrm {Dom}(f; u)$ . We shall now prove that

$$ \begin{align*}\textbf{V}^{(\mathbb{A})} \models \forall x (x \in u \wedge x \neq \varnothing \to \exists z \exists y (\mathrm{Pair}(z; x, y) \wedge z \in f \wedge y \in x))).\end{align*} $$

Consider an element $v \in \textbf{V}^{(\mathbb {A})}$ such that . Then, there exists an element $x \in \mathrm {dom}_D(u)$ such that $\textbf{V}^{(\mathbb {A})} \models v = x$ , and x is not 0-like. Consider the equivalence class $\bar {x}$ containing x in $\mathrm {Part}(\mathrm {dom}_D(u))$ . By the construction of f, there exists $w_{\bar {x}} \in \mathrm {dom}(f)$ which is an $(a_{\bar {x}}, b_{\bar {x}})$ -like element, where $a_{\bar {x}} \in \bar {x}$ and $b_{\bar {x}} \in \mathrm {dom}_D(a_{\bar {x}})$ . Since $a_{\bar {x}} \in \bar {x}$ , we get that $\textbf{V}^{(\mathbb {A})} \models a_{\bar {x}} = x$ , which implies $\textbf{V}^{(\mathbb {A})} \models a_{\bar {x}} = v$ . Hence, we can derive that $\textbf{V}^{(\mathbb {A})} \models \mathrm {Pair}(w_{\bar {x}}; v, b_{\bar {x}}) \wedge w_{\bar {x}} \in f \wedge b_{\bar {x}} \in v$ . Hence combining all the above results we can finally conclude that $\textbf{V}^{(\mathbb {A})} \models \mathrm {AC}$ .⊣

Proof. We shall prove (iii) only; statements (i) and (ii) can be proved similarly. Let $g : \mathrm {Part}(\mathrm {dom}_D(u)) \rightarrow \mathrm {Part}(\mathrm {dom}_D(v))$ be a bijection in $\textbf{V}$ . Let $\bar {x} \in \mathrm {Part}(\mathrm {dom}_D(u))$ be arbitrarily chosen and $ a \in \bar {x}$ and $b \in g(\bar {x})$ be arbitrarily fixed elements. Then, by definition, $a, b \in \textbf{V}^{(\mathbb {A})}$ . Consider two elements $p, q \in \textbf{V}^{(\mathbb {A})}$ , defined by $\mathrm {dom}(p) = \{a\}$ , $\mathrm {dom}(q) = \{a, b\}$ , and $\mathrm {ran}(p) = \{\textbf {1}\} = \mathrm {ran}(q)$ . Let us now take the element $w_{\bar {x}}$ of $\textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}(w_{\bar {x}}) = \{p, q\}$ and $\mathrm {ran}(w_{\bar {x}}) = \{ \textbf {1} \}$ . It can be proved that $w_{\bar {x}}$ is ( $a, b$ )-like. Clearly, for each $\bar {x} \in \mathrm {Part}(\mathrm {dom}_D(u))$ , we have one element $w_{\bar {x}}$ which is ( $a, b$ )-like for some $a \in \bar {x}$ and $b \in g(\bar {x})$ . Now consider $f \in \textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}(f) = \{w_{\bar {x}} : \bar {x} \in \mathrm {Part}(\mathrm {dom}_D(u))\}$ and $\mathrm {ran}(f) = \{ \textbf {1} \}$ ; such a function f exists in $\textbf{V}$ , and hence in $\textbf{V}^{(\mathbb {A})}$ , since $\textbf{V} \models \mathrm {AC}$ . By the definition, it is immediate that .

Conversely, let $\textbf{V}^{(\mathbb {A})} \models \exists f \; \mathrm {BijFunc}(f;u, v)$ . We shall prove that there exists a bijection between $\mathrm {Part}(\mathrm {dom}_D(u))$ and $\mathrm {Part}(\mathrm {dom}_D(v))$ . By the assumption, there exists $f \in \textbf{V}^{(\mathbb {A})}$ such that

By analysing all these, we conclude the following.

1. (i) If $w \in \mathrm {dom}_D(f)$ then w is ( $a, b$ )-like for some $a, b \in \textbf{V}^{(\mathbb {A})}$ .

2. (ii) For any $w, w' \in \mathrm {dom}_D(f)$ , if w is ( $a, b$ )-like and $w'$ is ( $c, d$ )-like for some $a, b, c, d \in \textbf{V}^{(\mathbb {A})}$ , where $a \sim c$ holds, then $b \sim d$ also holds.

3. (iii) For every $a \in \mathrm {dom}_D(u)$ there exist $w \in \mathrm {dom}_D(f)$ and $b \in \textbf{V}^{(\mathbb {A})}$ such that w is ( $a, b$ )-like.

4. (iv) For every $w \in \mathrm {dom}_D(f)$ there exist $a \in \mathrm {dom}_D(u)$ and $b \in \textbf{V}^{(\mathbb {A})}$ such that w is ( $a, b$ )-like.

5. (v) For every $w \in \mathrm {dom}_D(f)$ there exist $b \in \mathrm {dom}_D(v)$ and $a \in \textbf{V}^{(\mathbb {A})}$ such that w is ( $a, b$ )-like.

6. (vi) For any $w, w' \in \mathrm {dom}_D(f)$ and $a, b, c, d \in \textbf{V}^{(\mathbb {A})}$ if w is ( $a, b$ )-like, $w'$ is ( $c, d$ )-like, and $b \sim d$ , then $a \sim c$ .

7. (vii) If $b \in \mathrm {dom}_D(v)$ then there exist $a \in \mathrm {dom}_D(u)$ and $w \in \mathrm {dom}_D(f)$ such that w is ( $a, b$ )-like.

Combining all these results, we get a bijection between $\mathrm {Part}(\mathrm {dom}_D(u))$ and $\mathrm {Part}(\mathrm {dom}_D(v))$ in $\textbf{V}$ .⊣

Proof. Let the Well-Ordering Theorem hold in $\textbf{V}$ and hence $\textbf{V} \models \mathrm {AC}$ . Suppose $u \in \textbf{V}^{(\mathbb {A})}$ is any arbitrary element. Then, $\mathrm {Part}(\mathrm {dom}_D(u))$ is an element of $\textbf{V}$ . By our assumption, there exists $\alpha \in \mathrm {ORD}$ and a bijection between $\mathrm {Part}(\mathrm {dom}_D(u))$ and $\alpha $ in $\textbf{V}$ . Let us now consider any $\alpha $ -like element v. Then, by using Theorems 13 and 23(iii) we get that $\textbf{V}^{(\mathbb {A})} \models \mathsf {Ord}(v) \wedge \exists f~\mathrm {BijFunc}(f;u, v))$ . This completes the proof. ⊣

Observation 30. For any $u \in \textbf{V}^{(\mathbb {A})}$ , if $|\mathrm {Part}(\mathrm {dom}_D(u))|_{\textbf{V}} = \kappa $ , then using Corollary 24 and Theorem 28 it can be concluded that $\mathsf {Cardinal}_u$ consists of all $\kappa $ -like elements. This implies that $\textbf{V}^{(\mathbb {A})}$ cannot separate the elements in $\mathsf {Cardinal}_u$ , i.e., for any two cardinal numbers v and w of u, $\textbf{V}^{(\mathbb {A})} \models v = w$ . In particular, for any $v \in \mathsf {Cardinal}_u$ , $\textbf{V}^{(\mathbb {A})} \models v = \hat {\kappa }$ .

Proof. The proof can be done by using Observation 30 and the fact that, for any $u, v, w \in \textbf{V}^{(\mathbb {A})}$ , if both $\textbf{V}^{(\mathbb {A})} \models \exists f ~ \mathrm {BijFunc}(f;u, v)$ and $\textbf{V}^{(\mathbb {A})} \models \exists g ~ \mathrm {BijFunc}(g;v, w)$ hold then $\textbf{V}^{(\mathbb {A})} \models \exists h ~ \mathrm {BijFunc}(h;u, w)$ holds as well.⊣

Proof. A cardinal number of u is less than a cardinal number of v $\iff $ for any $p \in \mathsf {Cardinal}_u$ and $q \in \mathsf {Cardinal}_v$ there is an injection from $\mathrm {Part}(\mathrm {dom}_D(p))$ into $\mathrm {Part}(\mathrm {dom}_D(q))$ but no injection from $\mathrm {Part}(\mathrm {dom}_D(q))$ into $\mathrm {Part}(\mathrm {dom}_D(p))$ in $\textbf{V}$ ; this is because, applying Theorem 23, we get bijections between $\mathrm {Part}(\mathrm {dom}_D(u))$ and $\mathrm {Part}(\mathrm {dom}_D(p))$ and also between $\mathrm {Part}(\mathrm {dom}_D(v))$ and $\mathrm {Part}(\mathrm {dom}_D(q))$ in $\textbf{V}$ , $\iff \textbf{V}^{(\mathbb {A})} \models p \in q$ , since, by Observation 30, p is $\kappa $ -like and q is $\eta $ -like, $\iff \textbf{V}^{(\mathbb {A})} \models \hat {\kappa } \in \hat {\eta }$ .⊣

Proof. As an application of Theorem 23, there exist injections from $\mathrm {Part}(\mathrm {dom}_D(u))$ into $\mathrm {Part}(\mathrm {dom}_D(v))$ and from $\mathrm {Part}(\mathrm {dom}_D(v))$ into $\mathrm {Part}(\mathrm {dom}_D(u))$ in $\textbf{V}$ . Then, by the Schröder–Bernstein theorem in $\textbf{V}$ , there exists a bijection between $\mathrm {Part}(\mathrm {dom}_D(u))$ and $\mathrm {Part}(\mathrm {dom}_D(v))$ . Hence, one more application of Theorem 23 completes the proof.⊣

Proof. It is given that

. So, both

and

are in D, i.e.,

From the first condition, we get that if for any $x \in \textbf{V}^{(\mathbb {A})}$ ,

, i.e., for any $t \in \mathrm {dom}_D(x)$ there exists some $t' \in \mathrm {dom}_D(u)$ such that

, then

as well, i.e., there exists $x' \in \mathrm {dom}_D(v)$ such that

. From the second condition, it can be said that

i.e., for any $x \in \mathrm {dom}_D(v)$ and $t \in \mathrm {dom}_D(x)$ , there exists some $t' \in \mathrm {dom}_D(u)$ such that

.

By the given condition, $|\mathrm {Part}(\mathrm {dom}_D(u))|_{\textbf{V}} = \kappa $ . Hence, there exists a bijection f between $\mathrm {Part}(\mathrm {dom}_D(u))$ and $\kappa $ in $\textbf{V}$ . Let $A \subseteq _{\textbf{V}} \kappa $ be arbitrarily chosen. Then consider $\bigcup ~ f^{-1}(A)$ in $\textbf{V}$ . Clearly, $\bigcup ~ f^{-1}(A) \subseteq _{\textbf{V}} \mathrm {dom}_D(u)$ . Construct an element $x \in \textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}(x) = \bigcup ~ f^{-1}(A)$ and $\mathrm {ran}(x) = \{\textbf {1}\}$ . Therefore, we have . Hence, using the first condition, we can say that there exists an element $x_A \in \mathrm {dom}_D(v)$ such that .

Now suppose $x \in \mathrm {dom}_D(v)$ is an arbitrary element. We shall prove that there exists $A \subseteq _{\textbf{V}} \kappa $ such that . By the second condition, for any $t \in \mathrm {dom}_D(x)$ there exists an element $t' \in \mathrm {dom}_D(u)$ such that . Now consider the class $[t']$ in $\mathrm {Part}(\mathrm {dom}_D(u))$ and the image of it under the bijective function f. Let . Clearly, $A \subseteq _{\textbf{V}} \kappa $ . Construct an element $x_A \in \textbf{V}^{(\mathbb {A})}$ such that $\mathrm {dom}_D(x_A) = \bigcup ~f^{-1}(A)$ and $\mathrm {ran}(A) = \{\textbf {1}\}$ . Hence, by the construction we get .

Let us take $A, B \subseteq _{\textbf{V}}$ such that $A \neq B$ . It will be proved that . Without loss of generality, let there exist $a \in A$ where $a \notin B$ . By the construction, $\mathrm {dom}(x_A) = \bigcup ~ f^{-1}(A)$ and $\mathrm {dom}(x_B) = \bigcup ~ f^{-1}(B)$ , whereas $\mathrm {ran}(x_A) = \{\textbf {1}\} = \mathrm {ran}(x_B)$ . So for any $t \in f^{-1}(a)$ , $x_A(t) = \textbf {1}$ , whereas by the assumption, . Hence, we can conclude that .⊣

Proof. Let $\textbf{V} \models \mathrm {GCH}$ ; we shall prove that $\textbf{V}^{(\mathbb {A})} \models \mathrm {GCH}$ . Suppose $u \in \textbf{V}^{(\mathbb {A})}$ is an infinite element and $\hat {\kappa } \in \mathsf {Cardinal}_u$ . By this assumption we get that for any natural number in $\mathrm {ORD}$ there exists a subset of $\kappa $ which has the cardinality of that natural number in $\textbf{V}$ . Hence, $\kappa $ is an infinite cardinal in $\textbf{V}$ . Let v be a power set of u in $\textbf{V}^{(\mathbb {A})}$ . Then, by using Corollary 39, we get that $\hat {(2^\kappa )} \in \mathsf {Cardinal}_v$ . If possible, let there exist $w \in \textbf{V}^{(\mathbb {A})}$ such that $|u|_{\textbf{V}^{(\mathbb {A})}} < |w|_{\textbf{V}^{(\mathbb {A})}} < |v|_{\textbf{V}^{(\mathbb {A})}}$ . Let $\hat {\eta } \in \mathsf {Cardinal}_w$ . Hence, by Theorem 23, we get $\kappa < \eta < 2^\kappa $ in $\textbf{V}$ , which contradicts our assumption that $\textbf{V} \models \mathrm {GCH}$ .

Conversely, suppose $\textbf{V} \nvDash \mathrm {GCH}$ . Then in $\textbf{V}$ , there are infinite cardinals $\kappa $ and $\eta $ such that $\kappa < \eta < 2^\kappa $ . Now consider the element $\hat {\kappa } \in \textbf{V}^{(\mathbb {A})}$ . Clearly, by definition, $\hat {\kappa }$ is an infinite cardinal number in $\textbf{V}^{(\mathbb {A})}$ as well. Let v be a power set of $\hat {\kappa }$ in $\textbf{V}^{(\mathbb {A})}$ . Again by using Corollary 39, it can be concluded that ( $\hat {2^\kappa }$ ) $\in \mathsf {Cardinal}_v$ . Hence, by our assumption, $|\hat {\kappa }|_{\textbf{V}^{(\mathbb {A})}} < |\hat {\eta }|_{\textbf{V}^{(\mathbb {A})}} < |v|_{\textbf{V}^{(\mathbb {A})}}$ , i.e., $\textbf{V}^{(\mathbb {A})} \nvDash \mathrm {GCH}$ .⊣