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Theorem sb6x 2531
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypothesis
Ref Expression
sb6x.1 𝑥𝜑
Assertion
Ref Expression
sb6x ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3 𝑥𝜑
21sbf 2527 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 biidd 252 . . 3 (𝑥 = 𝑦 → (𝜑𝜑))
41, 3equsal 2446 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜑)
52, 4bitr4i 267 1 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wal 1629  wnf 1856  [wsb 2049
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-12 2203  ax-13 2408
This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-ex 1853  df-nf 1858  df-sb 2050
This theorem is referenced by: (None)
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