Theorem sb5f 1202

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

Hypothesis

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

Assertion

Ref	Expression
sb5f	|- ([y / x]ph <-> E.x(x = y /\ ph))

Proof of Theorem sb5f

Step	Hyp	Ref	Expression
1		equs45f.1	. . 3 |- (ph -> A.yph)
2	1	sb6f 1201	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
3	1	equs45f 1200	. 2 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))
4	2, 3	bitr4 176	1 |- ([y / x]ph <-> E.x(x = y /\ ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 954   = wceq 956  E.wex 980  [wsbc 1170

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-11 967  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123

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

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