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Theorem sbequ8 1743
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 = 𝑦𝜑))

Proof of Theorem sbequ8
StepHypRef Expression
1 pm5.4 242 . . 3 ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ↔ (𝑥 = 𝑦𝜑))
2 simpl 106 . . . . . 6 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) → 𝑥 = 𝑦)
3 pm3.35 333 . . . . . 6 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) → 𝜑)
42, 3jca 294 . . . . 5 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
5 simpl 106 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝑥 = 𝑦)
6 pm3.4 320 . . . . . 6 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
75, 6jca 294 . . . . 5 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)))
84, 7impbii 121 . . . 4 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) ↔ (𝑥 = 𝑦𝜑))
98exbii 1512 . . 3 (∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) ↔ ∃𝑥(𝑥 = 𝑦𝜑))
101, 9anbi12i 441 . 2 (((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑))) ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
11 df-sb 1662 . 2 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑))))
12 df-sb 1662 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
1310, 11, 123bitr4ri 206 1 ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wex 1397  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-4 1416  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by:  sbidm  1747
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