Theorem sbequ8 1768

Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.) (Proof revised by Jim Kingdon, 20-Jan-2018.)

Assertion

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

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		pm5.4 247	. . 3 ⊢ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ↔ (𝑥 = 𝑦 → 𝜑))
2		simpl 107	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)) → 𝑥 = 𝑦)
3		pm3.35 339	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)) → 𝜑)
4	2, 3	jca 300	. . . . 5 ⊢ ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)) → (𝑥 = 𝑦 ∧ 𝜑))
5		simpl 107	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝑥 = 𝑦)
6		pm3.4 326	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑))
7	5, 6	jca 300	. . . . 5 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)))
8	4, 7	impbii 124	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)) ↔ (𝑥 = 𝑦 ∧ 𝜑))
9	8	exbii 1536	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)) ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
10	1, 9	anbi12i 447	. 2 ⊢ (((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑))) ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
11		df-sb 1686	. 2 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑))))
12		df-sb 1686	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
13	10, 11, 12	3bitr4ri 211	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103  ∃wex 1421  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  sbidm  1772