Theorem sbequ12a 1166

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ12a	|- (x = y -> ([y / x]ph <-> [x / y]ph))

Proof of Theorem sbequ12a

Step	Hyp	Ref	Expression
1		sbequ12 1164	. 2 |- (x = y -> (ph <-> [y / x]ph))
2		sbequ12 1164	. . 3 |- (y = x -> (ph <-> [x / y]ph))
3	2	equcoms 1117	. 2 |- (x = y -> (ph <-> [x / y]ph))
4	1, 3	bitr3d 528	1 |- (x = y -> ([y / x]ph <-> [x / y]ph))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 1099  [wsbc 1153

This theorem is referenced by:  sbco3 1241

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-gen 955  ax-8 1101  ax-9 1102  ax-12 1104

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-sb 1155

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