Theorem sbequ8 2505

Description: Elimination of equality from antecedent after substitution. Usage of this theorem is discouraged because it depends on ax-13 2372. (Contributed by NM, 5-Aug-1993.) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018.) Revise df-sb 2068. (Revised by Wolf Lammen, 28-Jul-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
sbequ8	⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		equsb1 2495	. . 3 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
2	1	a1bi 363	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
3		sbim 2300	. 2 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
4	2, 3	bitr4i 277	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 205  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171  ax-13 2372

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1783  df-nf 1787  df-sb 2068

This theorem is referenced by: (None)