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Theorem bj-sseq 7185
 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 445 . 2 (φ → ((ψ χ) ↔ (AB BA)))
4 eqss 2937 . 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 1228   ⊆ wss 2894 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 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-11 1378  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004 This theorem depends on definitions:  df-bi 110  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-in 2901  df-ss 2908 This theorem is referenced by: (None)
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