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Theorem sbequ8 1772
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 pm5.4 247 . . 3 ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ↔ (𝑥 = 𝑦𝜑))
2 simpl 107 . . . . . 6 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) → 𝑥 = 𝑦)
3 pm3.35 339 . . . . . 6 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) → 𝜑)
42, 3jca 300 . . . . 5 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
5 simpl 107 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝑥 = 𝑦)
6 pm3.4 326 . . . . . 6 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
75, 6jca 300 . . . . 5 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)))
84, 7impbii 124 . . . 4 ((𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) ↔ (𝑥 = 𝑦𝜑))
98exbii 1539 . . 3 (∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑)) ↔ ∃𝑥(𝑥 = 𝑦𝜑))
101, 9anbi12i 448 . 2 (((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑))) ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
11 df-sb 1690 . 2 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ((𝑥 = 𝑦 → (𝑥 = 𝑦𝜑)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝑥 = 𝑦𝜑))))
12 df-sb 1690 . 2 ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
1310, 11, 123bitr4ri 211 1 ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥](𝑥 = 𝑦𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  wex 1424  [wsb 1689
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 1379  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-4 1443  ax-ial 1470
This theorem depends on definitions:  df-bi 115  df-sb 1690
This theorem is referenced by:  sbidm  1776
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