Theorem sb6f 1197

Description: Equivalence for substitution when y is not free in ph.

Hypothesis

Ref	Expression
equs45f.1	|- (ph -> A.yph)

Assertion

Ref	Expression
sb6f	|- ([y / x]ph <-> A.x(x = y -> ph))

Proof of Theorem sb6f

Step	Hyp	Ref	Expression
1		equs45f.1	. . . 4 |- (ph -> A.yph)
2	1	sbimi 1169	. . 3 |- ([y / x]ph -> [y / x]A.yph)
3		sb4a 1195	. . 3 |- ([y / x]A.yph -> A.x(x = y -> ph))
4	2, 3	syl 10	. 2 |- ([y / x]ph -> A.x(x = y -> ph))
5		sb2 1173	. 2 |- (A.x(x = y -> ph) -> [y / x]ph)
6	4, 5	impbi 157	1 |- ([y / x]ph <-> A.x(x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 951   = wceq 953  [wsbc 1166

This theorem is referenced by:  sb5f 1198

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 960  ax-11 964  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168

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