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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 = 𝑦𝜑))

Proof of Theorem sbequ8
StepHypRef Expression
1 pm5.4 376 . . . 4 ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ↔ (𝑥 = 𝑦𝜑))
21bicomi 213 . . 3 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)))
3 abai 832 . . . 4 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)))
43exbii 1764 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)))
52, 4anbi12i 729 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑))))
6 df-sb 1868 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
7 df-sb 1868 . 2 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑))))
85, 6, 73bitr4i 291 1 ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦𝜑))
 Colors of variables: wff setvar class 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)
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