Theorem sbequ12r 1219

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 1218	. 2 |- (y = x -> (ph <-> [x / y]ph))
2		equcom 1166	. 2 |- (x = y <-> y = x)
3		bicom 523	. 2 |- (([x / y]ph <-> ph) <-> (ph <-> [x / y]ph))
4	1, 2, 3	3imtr4i 217	1 |- (x = y -> ([x / y]ph <-> ph))

Syntax hints:   -> wi 3   <-> wb 144   = wceq 992  [wsbc 1207

This theorem is referenced by:  sb5rf 1297  tfindes 3215  findes 3248  isarep1 3683  axrepndlem1 5098  axrepndlem2 5099

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 999  ax-8 1000  ax-12 1004  ax-4 1009  ax-5o 1011  ax-6o 1014  ax-9o 1159

This theorem depends on definitions:  df-bi 145  df-an 223  df-ex 1017  df-sb 1209

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