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Theorem sb6x 1662
 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 (𝜑 → ∀𝑥𝜑)
Assertion
Ref Expression
sb6x ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3 (𝜑 → ∀𝑥𝜑)
21sbh 1659 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 biidd 161 . . 3 (𝑥 = 𝑦 → (𝜑𝜑))
41, 3equsalh 1614 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜑)
52, 4bitr4i 176 1 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 98  ∀wal 1241  [wsb 1645 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 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-i9 1423  ax-ial 1427 This theorem depends on definitions:  df-bi 110  df-sb 1646 This theorem is referenced by: (None)
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