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Theorem sbi1 1227
Description: Removal of implication from substitution.
Assertion
Ref Expression
sbi1 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))

Proof of Theorem sbi1
StepHypRef Expression
1 sbequ2 1175 . . . . 5 |- (x = y -> ([y / x](ph -> ps) -> (ph -> ps)))
2 sbequ2 1175 . . . . 5 |- (x = y -> ([y / x]ph -> ph))
31, 2syl5d 55 . . . 4 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> ps)))
4 sbequ1 1174 . . . 4 |- (x = y -> (ps -> [y / x]ps))
53, 4syl6d 56 . . 3 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
65a4s 981 . 2 |- (A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
7 sb4 1218 . . . 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)))
9819.20ii 992 . . . . 5 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> A.x(x = y -> ps)))
10 sb2 1173 . . . . 5 |- (A.x(x = y -> ps) -> [y / x]ps)
119, 10syl6 22 . . . 4 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> [y / x]ps))
127, 11syl6 22 . . 3 |- (-. A.x x = y -> ([y / x](ph -> ps) -> (A.x(x = y -> ph) -> [y / x]ps)))
13 sb4 1218 . . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x(x = y -> ph)))
1412, 13syl5d 55 . 2 |- (-. A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
156, 14pm2.61i 126 1 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))
Colors of variables: wff set class
Syntax hints:  -. wn 2   -> wi 3  A.wal 951  [wsbc 1166
This theorem is referenced by:  sbim 1229
This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-10 963  ax-12 965  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-11o 1213
This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168
Copyright terms: Public domain