Theorem sbequ8 1245

Description: Elimination of equality from antecedent after substitution.

Assertion

Ref	Expression
sbequ8	|- ([y / x]ph <-> [y / x](x = y -> ph))

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		equsb1 1191	. . 3 |- [y / x]x = y
2	1	a1bi 197	. 2 |- ([y / x]ph <-> ([y / x]x = y -> [y / x]ph))
3		sbim 1232	. 2 |- ([y / x](x = y -> ph) <-> ([y / x]x = y -> [y / x]ph))
4	2, 3	bitr4 176	1 |- ([y / x]ph <-> [y / x](x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 954  [wsbc 1168

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-10 964  ax-12 966  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170

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