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Theorem bj-sseq 10318
 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 450 . 2 (𝜑 → ((𝜓𝜒) ↔ (𝐴𝐵𝐵𝐴)))
4 eqss 2988 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
53, 4syl6bbr 191 1 (𝜑 → ((𝜓𝜒) ↔ 𝐴 = 𝐵))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 101   ↔ wb 102   = wceq 1259   ⊆ wss 2945 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-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-11 1413  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038 This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-in 2952  df-ss 2959 This theorem is referenced by: (None)
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