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 = 𝑦 → 𝜑))

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		equsb1 2451	. . 3 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
2	1	a1bi 355	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
3		sbim 2237	. 2 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
4	2, 3	bitr4i 270	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

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)