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Theorem trcleq2lem 13578
 Description: Equality implies bijection. (Contributed by RP, 5-May-2020.)
Assertion
Ref Expression
trcleq2lem (𝐴 = 𝐵 → ((𝑅𝐴 ∧ (𝐴𝐴) ⊆ 𝐴) ↔ (𝑅𝐵 ∧ (𝐵𝐵) ⊆ 𝐵)))

Proof of Theorem trcleq2lem
StepHypRef Expression
1 sseq2 3590 . 2 (𝐴 = 𝐵 → (𝑅𝐴𝑅𝐵))
2 id 22 . . . 4 (𝐴 = 𝐵𝐴 = 𝐵)
32, 2coeq12d 5208 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐵𝐵))
43, 2sseq12d 3597 . 2 (𝐴 = 𝐵 → ((𝐴𝐴) ⊆ 𝐴 ↔ (𝐵𝐵) ⊆ 𝐵))
51, 4anbi12d 743 1 (𝐴 = 𝐵 → ((𝑅𝐴 ∧ (𝐴𝐴) ⊆ 𝐴) ↔ (𝑅𝐵 ∧ (𝐵𝐵) ⊆ 𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383   = wceq 1475   ⊆ wss 3540   ∘ ccom 5042 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-in 3547  df-ss 3554  df-br 4584  df-opab 4644  df-co 5047 This theorem is referenced by:  cvbtrcl  13579  trcleq12lem  13580  trclublem  13582  cotrtrclfv  13601  trclun  13603  trclexi  36946  dftrcl3  37031
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