cbvsbc

Description: Change bound variables in a wff substitution. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker cbvsbcw when possible. (Contributed by Jeff Hankins, 19-Sep-2009) (Proof shortened by Andrew Salmon, 8-Jun-2011) (New usage is discouraged.)

	Ref	Expression
Hypotheses	cbvsbc.1	$⊢ Ⅎ y φ$
	cbvsbc.2	$⊢ Ⅎ x ψ$
	cbvsbc.3	$⊢ x = y \to (φ \leftrightarrow ψ)$
Assertion	cbvsbc	$⊢ \underset{˙}{[} A / x \underset{˙}{]} φ \leftrightarrow \underset{˙}{[} A / y \underset{˙}{]} ψ$

Proof

Step	Hyp	Ref	Expression
1		cbvsbc.1	$⊢ Ⅎ y φ$
2		cbvsbc.2	$⊢ Ⅎ x ψ$
3		cbvsbc.3	$⊢ x = y \to (φ \leftrightarrow ψ)$
4	1 2 3	cbvab	$⊢ \{x \| φ\} = \{y \| ψ\}$
5	4	eleq2i	$⊢ A \in \{x \| φ\} \leftrightarrow A \in \{y \| ψ\}$
6		df-sbc	$⊢ \underset{˙}{[} A / x \underset{˙}{]} φ \leftrightarrow A \in \{x \| φ\}$
7		df-sbc	$⊢ \underset{˙}{[} A / y \underset{˙}{]} ψ \leftrightarrow A \in \{y \| ψ\}$
8	5 6 7	3bitr4i	$⊢ \underset{˙}{[} A / x \underset{˙}{]} φ \leftrightarrow \underset{˙}{[} A / y \underset{˙}{]} ψ$

Metamath Proof Explorer

Theorem cbvsbc

Proof