Theorem sbequ12r 1178

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ12r	|- (x = y -> ([x / y]ph <-> ph))

Proof of Theorem sbequ12r

Step	Hyp	Ref	Expression
1		sbequ12 1177	. 2 |- (y = x -> (ph <-> [x / y]ph))
2		equcom 1125	. 2 |- (x = y <-> y = x)
3		bicom 518	. 2 |- (([x / y]ph <-> ph) <-> (ph <-> [x / y]ph))
4	1, 2, 3	3imtr4 219	1 |- (x = y -> ([x / y]ph <-> ph))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 953  [wsbc 1166

This theorem is referenced by:  sb5rf 1254  findes 3150  tfindes 3154  isarep1 3563  axrepndlem1 4916  axrepndlem2 4917

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 960  ax-8 961  ax-12 965  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168

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