Theorem sbequ12r 866

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 865	. 2 |- (y = x -> (ph <-> [x / y]ph))
2		eqcomb 812	. 2 |- (x = y <-> y = x)
3		bicom 398	. 2 |- (([x / y]ph <-> ph) <-> (ph <-> [x / y]ph))
4	1, 2, 3	3imtr4 192	1 |- (x = y -> ([x / y]ph <-> ph))

Syntax hints:   -> wi 2   <-> wb 127   = weq 797  [wsb 852

This theorem is referenced by:  findes 2400  tfindes 2404  axrepndlem1 3738  axrepndlem2 3739

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-gen 677  ax-8 798  ax-9 799  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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