Theorem sb6f 2039

Description: Equivalence for substitution when y is not free in ph. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6f.1	|- F/yph

Assertion

Ref	Expression
sb6f	|- ([y / x]ph <-> A.x(x = y -> ph))

Proof of Theorem sb6f

Step	Hyp	Ref	Expression
1		sb6f.1	. . . . 5 |- F/yph
2	1	nfri 1762	. . . 4 |- (ph -> A.yph)
3	2	sbimi 1652	. . 3 |- ([y / x]ph -> [y / x]A.yph)
4		sb4a 1923	. . 3 |- ([y / x]A.yph -> A.x(x = y -> ph))
5	3, 4	syl 15	. 2 |- ([y / x]ph -> A.x(x = y -> ph))
6		sb2 2023	. 2 |- (A.x(x = y -> ph) -> [y / x]ph)
7	5, 6	impbii 180	1 |- ([y / x]ph <-> A.x(x = y -> ph))

Syntax hints:   -> wi 4   <-> wb 176  A.wal 1540   F/wnf 1544  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sb5f  2040