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Theorem sbequ8 1724
 Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.) (Proof revised by Jim Kingdon, 20-Jan-2018.)
Assertion
Ref Expression
sbequ8 ([y / x]φ ↔ [y / x](x = yφ))

Proof of Theorem sbequ8
StepHypRef Expression
1 pm5.4 238 . . 3 ((x = y → (x = yφ)) ↔ (x = yφ))
2 simpl 102 . . . . . 6 ((x = y (x = yφ)) → x = y)
3 pm3.35 329 . . . . . 6 ((x = y (x = yφ)) → φ)
42, 3jca 290 . . . . 5 ((x = y (x = yφ)) → (x = y φ))
5 simpl 102 . . . . . 6 ((x = y φ) → x = y)
6 pm3.4 316 . . . . . 6 ((x = y φ) → (x = yφ))
75, 6jca 290 . . . . 5 ((x = y φ) → (x = y (x = yφ)))
84, 7impbii 117 . . . 4 ((x = y (x = yφ)) ↔ (x = y φ))
98exbii 1493 . . 3 (x(x = y (x = yφ)) ↔ x(x = y φ))
101, 9anbi12i 433 . 2 (((x = y → (x = yφ)) x(x = y (x = yφ))) ↔ ((x = yφ) x(x = y φ)))
11 df-sb 1643 . 2 ([y / x](x = yφ) ↔ ((x = y → (x = yφ)) x(x = y (x = yφ))))
12 df-sb 1643 . 2 ([y / x]φ ↔ ((x = yφ) x(x = y φ)))
1310, 11, 123bitr4ri 202 1 ([y / x]φ ↔ [y / x](x = yφ))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∃wex 1378  [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-ial 1424 This theorem depends on definitions:  df-bi 110  df-sb 1643 This theorem is referenced by:  sbidm  1728
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