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Theorem sbi2 1228
Description: Introduction of implication into substitution.
Assertion
Ref Expression
sbi2 |- (([y / x]ph -> [y / x]ps) -> [y / x](ph -> ps))

Proof of Theorem sbi2
StepHypRef Expression
1 sbn 1226 . . 3 |- ([y / x] -. ph <-> -. [y / x]ph)
2 pm2.21 76 . . . 4 |- (-. ph -> (ph -> ps))
32sbimi 1169 . . 3 |- ([y / x] -. ph -> [y / x](ph -> ps))
41, 3sylbir 201 . 2 |- (-. [y / x]ph -> [y / x](ph -> ps))
5 ax-1 4 . . 3 |- (ps -> (ph -> ps))
65sbimi 1169 . 2 |- ([y / x]ps -> [y / x](ph -> ps))
74, 6ja 137 1 |- (([y / x]ph -> [y / x]ps) -> [y / x](ph -> ps))
Colors of variables: wff set class
Syntax hints:  -. wn 2   -> wi 3  [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