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Theorem trcleq2lemRP 43620
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 5878 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐵𝐵))
32, 1sseq12d 4029 . 2 (𝐴 = 𝐵 → ((𝐴𝐴) ⊆ 𝐴 ↔ (𝐵𝐵) ⊆ 𝐵))
43cleq2lem 43598 1 (𝐴 = 𝐵 → ((𝑅𝐴 ∧ (𝐴𝐴) ⊆ 𝐴) ↔ (𝑅𝐵 ∧ (𝐵𝐵) ⊆ 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1537  wss 3963  ccom 5693
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ss 3980  df-br 5149  df-opab 5211  df-co 5698
This theorem is referenced by: (None)
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