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Theorem eqelb 38779
Description: Substitution of equal classes into element relation. (Contributed by Peter Mazsa, 17-Jul-2019.)
Assertion
Ref Expression
eqelb ((𝐴 = 𝐵𝐴𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))

Proof of Theorem eqelb
StepHypRef Expression
1 simpl 487 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵 = 𝐴)
2 eqeltr 38778 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵𝐶)
31, 2jca 520 . . 3 ((𝐵 = 𝐴𝐴𝐶) → (𝐵 = 𝐴𝐵𝐶))
4 eqcom 2776 . . . 4 (𝐵 = 𝐴𝐴 = 𝐵)
54anbi1i 635 . . 3 ((𝐵 = 𝐴𝐴𝐶) ↔ (𝐴 = 𝐵𝐴𝐶))
64anbi1i 635 . . 3 ((𝐵 = 𝐴𝐵𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
73, 5, 63imtr3i 294 . 2 ((𝐴 = 𝐵𝐴𝐶) → (𝐴 = 𝐵𝐵𝐶))
8 simpl 487 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴 = 𝐵)
9 eqeltr 38778 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴𝐶)
108, 9jca 520 . 2 ((𝐴 = 𝐵𝐵𝐶) → (𝐴 = 𝐵𝐴𝐶))
117, 10impbii 212 1 ((𝐴 = 𝐵𝐴𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 400   = wceq 1567  wcel 2149
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-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1807  df-cleq 2761  df-clel 2844
This theorem is referenced by:  eldmressnALTV  38817  raldmqseu  38903  inxpxrn  38956
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