Theorem sb6a 990

Description: Equivalence for substitution. Compare Theorem 6.2 of [Quine] p. 40.

Assertion

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

Distinct variable group(s):   x,y

Proof of Theorem sb6a

Step	Hyp	Ref	Expression
1		sb6 989	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sbequ12 865	. . . . 5 |- (y = x -> (ph <-> [x / y]ph))
3	2	eqcoms 813	. . . 4 |- (x = y -> (ph <-> [x / y]ph))
4	3	pm5.74i 443	. . 3 |- ((x = y -> ph) <-> (x = y -> [x / y]ph))
5	4	bial 695	. 2 |- (A.x(x = y -> ph) <-> A.x(x = y -> [x / y]ph))
6	1, 5	bitr 151	1 |- ([y / x]ph <-> A.x(x = y -> [x / y]ph))

Syntax hints:   -> wi 2   <-> wb 127  A.wal 672   = weq 797  [wsb 852

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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