Theorem sbequ12r 1184

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 1183	. 2 |- (y = x -> (ph <-> [x / y]ph))
2		equcom 1131	. 2 |- (x = y <-> y = x)
3		bicom 522	. 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 958  [wsbc 1172

This theorem is referenced by:  sb5rf 1261  findes 3166  tfindes 3170  isarep1 3583  axrepndlem1 4956  axrepndlem2 4957

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 965  ax-8 966  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174

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