Theorem sb6a 1339

Description: Equivalence for substitution.

Assertion

Ref	Expression
sb6a	|- ([y / x]ph <-> A.x(x = y -> [x / y]ph))

Distinct variable group:   x,y

Proof of Theorem sb6a

Step	Hyp	Ref	Expression
1		sb6 1269	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sbequ12 1183	. . . . 5 |- (y = x -> (ph <-> [x / y]ph))
3	2	equcoms 1132	. . . 4 |- (x = y -> (ph <-> [x / y]ph))
4	3	pm5.74i 586	. . 3 |- ((x = y -> ph) <-> (x = y -> [x / y]ph))
5	4	albii 1001	. 2 |- (A.x(x = y -> ph) <-> A.x(x = y -> [x / y]ph))
6	1, 5	bitr 173	1 |- ([y / x]ph <-> A.x(x = y -> [x / y]ph))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 956   = wceq 958  [wsbc 1172

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 965  ax-8 966  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-16 1212  ax-11o 1220

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174

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