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Theorem sb6x 1678
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 1675 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 biidd 165 . . 3 (𝑥 = 𝑦 → (𝜑𝜑))
41, 3equsalh 1630 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜑)
52, 4bitr4i 180 1 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 102  wal 1257  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-4 1416  ax-i9 1439  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by: (None)
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