Theorem sbequ8 1885

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.)

Assertion

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

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		pm5.4 377	. . . 4 ⊢ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ↔ (𝑥 = 𝑦 → 𝜑))
2	1	bicomi 214	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)))
3		abai 836	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)))
4	3	exbii 1774	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)))
5	2, 4	anbi12i 733	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑))))
6		df-sb 1881	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
7		df-sb 1881	. 2 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑))))
8	5, 6, 7	3bitr4i 292	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384  ∃wex 1704  [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-sb 1881

This theorem is referenced by: (None)