Theorem sbi1 2063

Description: Removal of implication from substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbi1	|- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))

Proof of Theorem sbi1

Step	Hyp	Ref	Expression
1		sbequ2 1650	. . . . 5 |- (x = y -> ([y / x]ph -> ph))
2		sbequ2 1650	. . . . 5 |- (x = y -> ([y / x](ph -> ps) -> (ph -> ps)))
3	1, 2	syl5d 62	. . . 4 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> ps)))
4		sbequ1 1918	. . . 4 |- (x = y -> (ps -> [y / x]ps))
5	3, 4	syl6d 64	. . 3 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
6	5	sps 1754	. 2 |- (A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
7		sb4 2053	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x(x = y -> ph)))
8		sb4 2053	. . . 4 |- (-. A.x x = y -> ([y / x](ph -> ps) -> A.x(x = y -> (ph -> ps))))
9		ax-2 7	. . . . . 6 |- ((x = y -> (ph -> ps)) -> ((x = y -> ph) -> (x = y -> ps)))
10	9	al2imi 1561	. . . . 5 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> A.x(x = y -> ps)))
11		sb2 2023	. . . . 5 |- (A.x(x = y -> ps) -> [y / x]ps)
12	10, 11	syl6 29	. . . 4 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> [y / x]ps))
13	8, 12	syl6 29	. . 3 |- (-. A.x x = y -> ([y / x](ph -> ps) -> (A.x(x = y -> ph) -> [y / x]ps)))
14	7, 13	syl5d 62	. 2 |- (-. A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
15	6, 14	pm2.61i 156	1 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))

Syntax hints:  -. wn 3   -> wi 4  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:  sbim  2065  spsbim  2076