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Theorem bj-sseq 9200
 Description: If two converse inclusions are characterized each by a formula, then equality is characterized by the conjunction of these formulas. (Contributed by BJ, 30-Nov-2019.)
Hypotheses
Ref Expression
bj-sseq.1 (φ → (ψAB))
bj-sseq.2 (φ → (χBA))
Assertion
Ref Expression
bj-sseq (φ → ((ψ χ) ↔ A = B))

Proof of Theorem bj-sseq
StepHypRef Expression
1 bj-sseq.1 . . 3 (φ → (ψAB))
2 bj-sseq.2 . . 3 (φ → (χBA))
31, 2anbi12d 442 . 2 (φ → ((ψ χ) ↔ (AB BA)))
4 eqss 2954 . 2 (A = B ↔ (AB BA))
53, 4syl6bbr 187 1 (φ → ((ψ χ) ↔ A = B))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242   ⊆ wss 2911 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-in 2918  df-ss 2925 This theorem is referenced by: (None)
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