HomeHome Metamath Proof Explorer < Previous   Next >
Related theorems
Unicode version

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
StepHypRef Expression
1 sbequ2 1823 . . . . 5 |- (x = y -> ([y / x]ph -> ph))
2 sbequ2 1823 . . . . 5 |- (x = y -> ([y / x](ph -> ps) -> (ph -> ps)))
31, 2syl5d 49 . . . 4 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> ps)))
4 sbequ1 1822 . . . 4 |- (x = y -> (ps -> [y / x]ps))
53, 4syl6d 54 . . 3 |- (x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
65a4s 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)))
109al2imi 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)
1210, 11syl6 42 . . . 4 |- (A.x(x = y -> (ph -> ps)) -> (A.x(x = y -> ph) -> [y / x]ps))
138, 12syl6 42 . . 3 |- (-. A.x x = y -> ([y / x](ph -> ps) -> (A.x(x = y -> ph) -> [y / x]ps)))
147, 13syl5d 49 . 2 |- (-. A.x x = y -> ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps)))
156, 14pm2.61i 192 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 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
Copyright terms: Public domain