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Theorem eqtrb 32761
Description: A transposition of equality. (Contributed by Thierry Arnoux, 20-Aug-2023.)
Assertion
Ref Expression
eqtrb ((𝐴 = 𝐵𝐴 = 𝐶) ↔ (𝐴 = 𝐵𝐵 = 𝐶))

Proof of Theorem eqtrb
StepHypRef Expression
1 simpl 487 . . 3 ((𝐴 = 𝐵𝐴 = 𝐶) → 𝐴 = 𝐵)
2 eqtr2 2790 . . 3 ((𝐴 = 𝐵𝐴 = 𝐶) → 𝐵 = 𝐶)
31, 2jca 520 . 2 ((𝐴 = 𝐵𝐴 = 𝐶) → (𝐴 = 𝐵𝐵 = 𝐶))
4 simpl 487 . . 3 ((𝐴 = 𝐵𝐵 = 𝐶) → 𝐴 = 𝐵)
5 eqtr 2789 . . 3 ((𝐴 = 𝐵𝐵 = 𝐶) → 𝐴 = 𝐶)
64, 5jca 520 . 2 ((𝐴 = 𝐵𝐵 = 𝐶) → (𝐴 = 𝐵𝐴 = 𝐶))
73, 6impbii 212 1 ((𝐴 = 𝐵𝐴 = 𝐶) ↔ (𝐴 = 𝐵𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 400   = wceq 1567
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1807  df-cleq 2761
This theorem is referenced by: (None)
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