Theorem sbequ12a 1183

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 1181	. 2 |- (x = y -> (ph <-> [y / x]ph))
2		sbequ12 1181	. . 3 |- (y = x -> (ph <-> [x / y]ph))
3	2	equcoms 1130	. 2 |- (x = y -> (ph <-> [x / y]ph))
4	1, 3	bitr3d 530	1 |- (x = y -> ([y / x]ph <-> [x / y]ph))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 956  [wsbc 1170

This theorem is referenced by:  sbco3 1257

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-8 964  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172

Copyright terms: Public domain