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Theorem eqelb 34002
Description: Substitution of equal classes into elementhood relation. (Contributed by Peter Mazsa, 17-Jul-2019.)
Assertion
Ref Expression
eqelb ((𝐴 = 𝐵𝐴𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))

Proof of Theorem eqelb
StepHypRef Expression
1 simpl 473 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵 = 𝐴)
2 eqeltr 34001 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵𝐶)
31, 2jca 554 . . 3 ((𝐵 = 𝐴𝐴𝐶) → (𝐵 = 𝐴𝐵𝐶))
4 eqcom 2629 . . . 4 (𝐵 = 𝐴𝐴 = 𝐵)
54anbi1i 731 . . 3 ((𝐵 = 𝐴𝐴𝐶) ↔ (𝐴 = 𝐵𝐴𝐶))
64anbi1i 731 . . 3 ((𝐵 = 𝐴𝐵𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
73, 5, 63imtr3i 280 . 2 ((𝐴 = 𝐵𝐴𝐶) → (𝐴 = 𝐵𝐵𝐶))
8 simpl 473 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴 = 𝐵)
9 eqeltr 34001 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴𝐶)
108, 9jca 554 . 2 ((𝐴 = 𝐵𝐵𝐶) → (𝐴 = 𝐵𝐴𝐶))
117, 10impbii 199 1 ((𝐴 = 𝐵𝐴𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 196  wa 384   = wceq 1483  wcel 1990
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-ext 2602
This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-cleq 2615  df-clel 2618
This theorem is referenced by: (None)
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