Theorem sb6a 2116

Description: Equivalence for substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Distinct variable group:   x,y

Allowed substitution hints:   ph(x,y)

Proof of Theorem sb6a

Step	Hyp	Ref	Expression
1		sb6 2099	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sbequ12 1919	. . . . 5 |- (y = x -> (ph <-> [x / y]ph))
3	2	equcoms 1681	. . . 4 |- (x = y -> (ph <-> [x / y]ph))
4	3	pm5.74i 236	. . 3 |- ((x = y -> ph) <-> (x = y -> [x / y]ph))
5	4	albii 1566	. 2 |- (A.x(x = y -> ph) <-> A.x(x = y -> [x / y]ph))
6	1, 5	bitri 240	1 |- ([y / x]ph <-> A.x(x = y -> [x / y]ph))

Syntax hints:   -> wi 4   <-> wb 176  A.wal 1540  [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)