a3
Proverバグ？解消のため次を準備しておきます。set\({\tt n}\).x【\(v \in \{ x_{1} , \cdots , x_{\tt n} \} \Rightarrow P \Longleftrightarrow ( v = x_{1} \Rightarrow P ) , \cdots , ( v = x_{\tt n} \Rightarrow P )\)】
a4
Cap.【\(x \in \bigcap \mathcal{X} \Longleftrightarrow \begin{cases} \exists X \, X \in \mathcal{X} \Rightarrow \forall X \in \mathcal{X} . x \in X \\ \not\exists X \, X \in \mathcal{X} \Rightarrow x \in \bigcap \mathcal{X} \end{cases}\)】\(\blacktriangleleft\)\V1.【\(\mathbb{V}_1 = \{ x \mid \not\exists y \, ( x \in y \in x ) \}\)】\(\blacktriangleleft\)
ax_r【\(\forall X ( \exists x \, x \in X \Rightarrow \exists Y \in X . \not\exists x \, ( x \in X , x \in Y ) )\)】\(\blacktriangleleft\)
b2
dom!.【\(\text{dom!} ( f ) = \{ x \mid \exists! y \, \langle x , y \rangle \in f \}\)】\(\blacktriangleleft\)b3
mapon.【\(\text{mapon} ( X ) = \{ f \in \text{Rel} \mid \text{dom} ( f ) \subset X \subset \text{dom!} ( f ) \}\)】\(\blacktriangleleft\)->..【\(X \to Y = \{ f \in \text{mapon} ( X ) \mid \forall x \in X . f ( x ) \in Y \}\)】\(\blacktriangleleft\)
mto.【\(p \in x \mathop{{\cdot}{\to}} y \Longleftrightarrow p = \langle x , y \rangle\)】\(\blacktriangleleft\)
->I.【\(X \stackrel{\rm I}\to Y = \{ f \in X \to Y \mid \forall y \, ! x \, \langle x , y \rangle \in f \}\)】\(\blacktriangleleft\)
->S.【\(X \stackrel{\rm S}\to Y = \{ f \in X \to Y \mid \forall y \in Y . \exists x \, \langle x , y \rangle \in f \}\)】\(\blacktriangleleft\)
->S.''【\(X \stackrel{\rm S}\to Y = \{ f \in \text{mapon} ( X ) \mid \text{im} ( f ) = Y \}\)】\(\blacktriangleleft\)
b4
=#.【\(X \stackrel{\#}= Y \Longleftrightarrow \exists f \, f \in X \stackrel{\rm IS}\to Y\)】\(\blacktriangleleft\)le#.【\(X \stackrel{\#}\le Y \Longleftrightarrow \exists f \, f \in X \stackrel{\rm I}\to Y\)】\(\blacktriangleleft\)
c1
0.【\(x \in 0 \Longleftrightarrow {\perp}\)】\(\blacktriangleleft\)suc.【\(x \in n \textit{+1} \Longleftrightarrow x \in n \mathbin{\rm o\!r} x = n\)】\(\blacktriangleleft\)
Ind.【\(\text{Ind} = \{ M \mid 0 \in M , \forall m \in M . m \textit{+1} \in M \}\)】\(\blacktriangleleft\)
\Mi1【\(\forall n \in \mathbb{M} . n \textit{+1} \in \mathbb{M}\)】\(\blacktriangleleft\)
<\Mi【\(n \in \mathbb{M} \Longrightarrow n \subset \mathbb{M}\)】\(\blacktriangleleft\)
\(\mathbb{M}\)が推移的であることを示します。機械は細い論理も平気で通します！
<\Ml【\(m \in n \in \mathbb{M} \Longrightarrow m \subsetneq n\)】\(\blacktriangleleft\)
<\Mr【\( m , n \in \mathbb{M} , m \subsetneq n \Longrightarrow m \in n\)】\(\blacktriangleleft\)
\({\tt n}\).【\(x \in {\tt n} \Longleftrightarrow x = 0 \mathbin{\rm o\!r} x = 1 \mathbin{\rm o\!r} \cdots \mathbin{\rm o\!r} x = {\tt n\,{\text -}\,1}\)】\(\blacktriangleleft\)
c2
+\M0【\(m \in \mathbb{M} , n \in \mathbb{M} \Longrightarrow m + n \in \mathbb{M}\)】\(\blacktriangleleft\)*\M0【\(m \in \mathbb{M} , n \in \mathbb{M} \Longrightarrow m \cdot n \in \mathbb{M}\)】\(\blacktriangleleft\)
b4
mapon:c.【\(\text{mapon:c} ( \mathcal{X} ) = \{ f \in \mathcal{X} \to \bigcup \mathcal{X} \mid \forall X , x \, ( \langle X , x \rangle \in f \Rightarrow x \in X ) \}\)】\(\blacktriangleleft\)dp.【\(x \in \Pi X \Longleftrightarrow \begin{cases} X \in \text{Map} \Longrightarrow x \in \text{mapon} ( \text{dom} ( X ) ) , \forall \lambda \in \text{dom} ( X ) . x _ \lambda \in X _ \lambda \\ X \notin \text{Map} \Longrightarrow x \in \Pi X \end{cases}\)】\(\blacktriangleleft\)