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Theorem sbequ8 2536
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 2061. (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 2523 . . 3 [𝑦 / 𝑥]𝑥 = 𝑦
21a1bi 364 . 2 ([𝑦 / 𝑥]𝜑 ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
3 sbim 2302 . 2 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
42, 3bitr4i 279 1 ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  [wsb 2060
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-10 2136  ax-12 2167  ax-13 2381
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-ex 1772  df-nf 1776  df-sb 2061
This theorem is referenced by: (None)
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