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Theorem trcleq2lemRP 38886
 Description: Equality implies bijection. (Contributed by RP, 5-May-2020.) (Proof modification is discouraged.)
Assertion
Ref Expression
trcleq2lemRP (𝐴 = 𝐵 → ((𝑅𝐴 ∧ (𝐴𝐴) ⊆ 𝐴) ↔ (𝑅𝐵 ∧ (𝐵𝐵) ⊆ 𝐵)))

Proof of Theorem trcleq2lemRP
StepHypRef Expression
1 id 22 . . . 4 (𝐴 = 𝐵𝐴 = 𝐵)
21, 1coeq12d 5532 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐵𝐵))
32, 1sseq12d 3852 . 2 (𝐴 = 𝐵 → ((𝐴𝐴) ⊆ 𝐴 ↔ (𝐵𝐵) ⊆ 𝐵))
43cleq2lem 38863 1 (𝐴 = 𝐵 → ((𝑅𝐴 ∧ (𝐴𝐴) ⊆ 𝐴) ↔ (𝑅𝐵 ∧ (𝐵𝐵) ⊆ 𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386   = wceq 1601   ⊆ wss 3791   ∘ ccom 5359 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2054  ax-9 2115  ax-10 2134  ax-11 2149  ax-12 2162  ax-ext 2753 This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2763  df-cleq 2769  df-clel 2773  df-nfc 2920  df-in 3798  df-ss 3805  df-br 4887  df-opab 4949  df-co 5364 This theorem is referenced by: (None)
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