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Theorem sb6x 1659
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)
Hypothesis
Ref Expression
sb6x.1 (φxφ)
Assertion
Ref Expression
sb6x ([y / x]φx(x = yφ))

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3 (φxφ)
21sbh 1656 . 2 ([y / x]φφ)
3 biidd 161 . . 3 (x = y → (φφ))
41, 3equsalh 1611 . 2 (x(x = yφ) ↔ φ)
52, 4bitr4i 176 1 ([y / x]φx(x = yφ))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98  wal 1240  [wsb 1642
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-4 1397  ax-i9 1420  ax-ial 1424
This theorem depends on definitions:  df-bi 110  df-sb 1643
This theorem is referenced by: (None)
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