Theorem sbi1 1216

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 1162	. . . . 5 |- (x = y -> ([y / x](ph -> ps) -> (ph -> ps)))
2		sbequ2 1162	. . . . 5 |- (x = y -> ([y / x]ph -> ph))
3	1, 2	syl5d 55	. . . 4 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> ps)))
4		sbequ1 1161	. . . 4 |- (x = y -> (ps -> [y / x]ps))
5	3, 4	syl6d 56	. . 3 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
6	5	a4s 960	. 2 |- (A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
7		sb4 1207	. . . 4 |- (-. A.x x = y -> ([y / x](ph -> ps) -> A.x(x = y -> (ph -> ps))))
8		ax-2 5	. . . . . 6 |- ((x = y -> (ph -> ps)) -> ((x = y -> ph) -> (x = y -> ps)))
9	8	19.20ii 971	. . . . 5 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> A.x(x = y -> ps)))
10		sb2 1160	. . . . 5 |- (A.x(x = y -> ps) -> [y / x]ps)
11	9, 10	syl6 22	. . . 4 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> [y / x]ps))
12	7, 11	syl6 22	. . 3 |- (-. A.x x = y -> ([y / x](ph -> ps) -> (A.x(x = y -> ph) -> [y / x]ps)))
13		sb4 1207	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x(x = y -> ph)))
14	12, 13	syl5d 55	. 2 |- (-. A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
15	6, 14	pm2.61i 126	1 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))

Syntax hints:  -. wn 2   -> wi 3  A.wal 950  [wsbc 1153

This theorem is referenced by:  sbim 1218

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-7 954  ax-gen 955  ax-8 1101  ax-9 1102  ax-10 1103  ax-12 1104  ax-11o 1202

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-sb 1155

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