 Mathbox for BJ < Previous   Next > Nearby theorems Mirrors  >  Home  >  ILE Home  >  Th. List  >   Mathboxes  >  bj-sseq GIF version

Theorem bj-sseq 11161
 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 457 . 2 (𝜑 → ((𝜓𝜒) ↔ (𝐴𝐵𝐵𝐴)))
4 eqss 3029 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
53, 4syl6bbr 196 1 (𝜑 → ((𝜓𝜒) ↔ 𝐴 = 𝐵))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1287   ⊆ wss 2988 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-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-11 1440  ax-4 1443  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067 This theorem depends on definitions:  df-bi 115  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-clel 2081  df-in 2994  df-ss 3001 This theorem is referenced by: (None)
 Copyright terms: Public domain W3C validator