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Theorem sbequ8 2467
 Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018.) Revise df-sb 2016. (Revised by Wolf Lammen, 28-Jul-2023.)
Assertion
Ref Expression
sbequ8 ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦𝜑))

Proof of Theorem sbequ8
StepHypRef Expression
1 equsb1 2451 . . 3 [𝑦 / 𝑥]𝑥 = 𝑦
21a1bi 355 . 2 ([𝑦 / 𝑥]𝜑 ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
3 sbim 2237 . 2 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
42, 3bitr4i 270 1 ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦𝜑))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198  [wsb 2015 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-10 2079  ax-12 2106  ax-13 2301 This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-ex 1743  df-nf 1747  df-sb 2016 This theorem is referenced by: (None)
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