Theorem sbequ8 2536

Description: Elimination of equality from antecedent after substitution. Usage of this theorem is discouraged because it depends on ax-13 2407. (Contributed by NM, 5-Aug-1993.) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018.) Revise df-sb 2100. (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 2526	. . 3 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
2	1	a1bi 365	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
3		sbim 2341	. 2 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
4	2, 3	bitr4i 281	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216  ax-13 2407

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by: (None)