Theorem sbequ8 1872

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		pm5.4 376	. . . 4 ⊢ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ↔ (𝑥 = 𝑦 → 𝜑))
2	1	bicomi 213	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)))
3		abai 832	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)))
4	3	exbii 1764	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑)))
5	2, 4	anbi12i 729	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑))))
6		df-sb 1868	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
7		df-sb 1868	. 2 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦 → 𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦 → 𝜑))))
8	5, 6, 7	3bitr4i 291	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∃wex 1695  [wsb 1867

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-sb 1868

This theorem is referenced by: (None)