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Theorem sb6x 2488
Description: Equivalence involving substitution for a variable not free. Usage of this theorem is discouraged because it depends on ax-13 2391. Usage of sb6 2093 is preferred, which requires fewer axioms. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.) (New usage is discouraged.)
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 2272 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 biidd 265 . . 3 (𝑥 = 𝑦 → (𝜑𝜑))
41, 3equsal 2440 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜑)
52, 4bitr4i 281 1 ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wal 1536  wnf 1785  [wsb 2069
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-12 2178  ax-13 2391
This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-nf 1786  df-sb 2070
This theorem is referenced by: (None)
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