Theorem sbequ8 1247

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 1193	. . 3 |- [y / x]x = y
2	1	a1bi 197	. 2 |- ([y / x]ph <-> ([y / x]x = y -> [y / x]ph))
3		sbim 1234	. 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 956  [wsbc 1170

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-10 966  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172

Copyright terms: Public domain