Theorem sb6a 1336

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 1266	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sbequ12 1180	. . . . 5 |- (y = x -> (ph <-> [x / y]ph))
3	2	equcoms 1129	. . . 4 |- (x = y -> (ph <-> [x / y]ph))
4	3	pm5.74i 583	. . 3 |- ((x = y -> ph) <-> (x = y -> [x / y]ph))
5	4	albii 998	. 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 953   = wceq 955  [wsbc 1169

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 962  ax-8 963  ax-12 967  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-16 1209  ax-11o 1217

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980  df-sb 1171

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