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Theorem trcleq2lem 14885
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 3974 . 2 (𝐴 = 𝐵 → (𝑅𝐴𝑅𝐵))
2 id 22 . . . 4 (𝐴 = 𝐵𝐴 = 𝐵)
32, 2coeq12d 5824 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐵𝐵))
43, 2sseq12d 3981 . 2 (𝐴 = 𝐵 → ((𝐴𝐴) ⊆ 𝐴 ↔ (𝐵𝐵) ⊆ 𝐵))
51, 4anbi12d 632 1 (𝐴 = 𝐵 → ((𝑅𝐴 ∧ (𝐴𝐴) ⊆ 𝐴) ↔ (𝑅𝐵 ∧ (𝐵𝐵) ⊆ 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 397   = wceq 1542  wss 3914  ccom 5641
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3449  df-in 3921  df-ss 3931  df-br 5110  df-opab 5172  df-co 5646
This theorem is referenced by:  cvbtrcl  14886  trcleq12lem  14887  trclublem  14889  cotrtrclfv  14906  trclun  14908  trclexi  41984  dftrcl3  42084
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