Theorem sbi1 1878

Description: Removal of implication from substitution.

Assertion

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

Proof of Theorem sbi1

Step	Hyp	Ref	Expression
1		sbequ2 1823	. . . . 5 |- (x = y -> ([y / x]ph -> ph))
2		sbequ2 1823	. . . . 5 |- (x = y -> ([y / x](ph -> ps) -> (ph -> ps)))
3	1, 2	syl5d 49	. . . 4 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> ps)))
4		sbequ1 1822	. . . 4 |- (x = y -> (ps -> [y / x]ps))
5	3, 4	syl6d 54	. . 3 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
6	5	a4s 1619	. 2 |- (A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
7		sb4 1869	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x(x = y -> ph)))
8		sb4 1869	. . . 4 |- (-. A.x x = y -> ([y / x](ph -> ps) -> A.x(x = y -> (ph -> ps))))
9		ax-2 5	. . . . . 6 |- ((x = y -> (ph -> ps)) -> ((x = y -> ph) -> (x = y -> ps)))
10	9	al2imi 1630	. . . . 5 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> A.x(x = y -> ps)))
11		sb2 1821	. . . . 5 |- (A.x(x = y -> ps) -> [y / x]ps)
12	10, 11	syl6 42	. . . 4 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> [y / x]ps))
13	8, 12	syl6 42	. . 3 |- (-. A.x x = y -> ([y / x](ph -> ps) -> (A.x(x = y -> ph) -> [y / x]ps)))
14	7, 13	syl5d 49	. 2 |- (-. A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
15	6, 14	pm2.61i 192	1 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))

Syntax hints:  -. wn 2   -> wi 3  A.wal 1584  [wsbc 1814

This theorem is referenced by:  sbim 1880  a4sbim 1890

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1592  ax-gen 1593  ax-10 1596  ax-12 1598  ax-4 1608  ax-5o 1610  ax-6o 1613  ax-9o 1763  ax-10o 1781  ax-11o 1864

This theorem depends on definitions:  df-bi 220  df-an 339  df-ex 1616  df-sb 1816

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