Theorem equs45f 1200

Description: Two ways of expressing substitution when y is not free in ph.

Hypothesis

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

Assertion

Ref	Expression
equs45f	|- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Proof of Theorem equs45f

Step	Hyp	Ref	Expression
1		equs45f.1	. . . . 5 |- (ph -> A.yph)
2	1	anim2i 335	. . . 4 |- ((x = y /\ ph) -> (x = y /\ A.yph))
3	2	19.22i 1039	. . 3 |- (E.x(x = y /\ ph) -> E.x(x = y /\ A.yph))
4		equs5a 1197	. . 3 |- (E.x(x = y /\ A.yph) -> A.x(x = y -> ph))
5	3, 4	syl 10	. 2 |- (E.x(x = y /\ ph) -> A.x(x = y -> ph))
6		equs4 1149	. 2 |- (A.x(x = y -> ph) -> E.x(x = y /\ ph))
7	5, 6	impbi 157	1 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 953   = wceq 955  E.wex 979

This theorem is referenced by:  sb5f 1202

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 962  ax-11 966  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980

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