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Theorem sbequ1 1695
 Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ1 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))

Proof of Theorem sbequ1
StepHypRef Expression
1 pm3.4 326 . . 3 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
2 19.8a 1525 . . 3 ((𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
3 df-sb 1690 . . 3 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
41, 2, 3sylanbrc 408 . 2 ((𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
54ex 113 1 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
 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  ax-ia2 105  ax-ia3 106  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-4 1443 This theorem depends on definitions:  df-bi 115  df-sb 1690 This theorem is referenced by:  sbequ12  1698  sbequi  1764  sb6rf  1778  mo2n  1973  bj-bdfindes  11289  bj-findes  11321
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