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Theorem sbequ2 1696
Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ2 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))

Proof of Theorem sbequ2
StepHypRef Expression
1 df-sb 1690 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2 simpl 107 . . 3 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
32com12 30 . 2 (𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → 𝜑))
41, 3syl5bi 150 1 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wex 1424  [wsb 1689
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104
This theorem depends on definitions:  df-bi 115  df-sb 1690
This theorem is referenced by:  stdpc7  1697  sbequ12  1698  sbequi  1764  mo23  1986  mopick  2023
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