Theorem sb6x 2029

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6x.1	|- F/xph

Assertion

Ref	Expression
sb6x	|- ([y / x]ph <-> A.x(x = y -> ph))

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 |- F/xph
2	1	sbf 2026	. 2 |- ([y / x]ph <-> ph)
3		biidd 228	. . 3 |- (x = y -> (ph <-> ph))
4	1, 3	equsal 1960	. 2 |- (A.x(x = y -> ph) <-> ph)
5	2, 4	bitr4i 243	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: (None)