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Theorem bj-sseq 14583
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 → ((𝜓𝜒) ↔ 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 473 . 2 (𝜑 → ((𝜓𝜒) ↔ (𝐴𝐵𝐵𝐴)))
4 eqss 3172 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
53, 4bitr4di 198 1 (𝜑 → ((𝜓𝜒) ↔ 𝐴 = 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105   = wceq 1353  wss 3131
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-in 3137  df-ss 3144
This theorem is referenced by: (None)
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