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Theorem eqelb 38740
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 486 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵 = 𝐴)
2 eqeltr 38739 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵𝐶)
31, 2jca 519 . . 3 ((𝐵 = 𝐴𝐴𝐶) → (𝐵 = 𝐴𝐵𝐶))
4 eqcom 2769 . . . 4 (𝐵 = 𝐴𝐴 = 𝐵)
54anbi1i 633 . . 3 ((𝐵 = 𝐴𝐴𝐶) ↔ (𝐴 = 𝐵𝐴𝐶))
64anbi1i 633 . . 3 ((𝐵 = 𝐴𝐵𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
73, 5, 63imtr3i 293 . 2 ((𝐴 = 𝐵𝐴𝐶) → (𝐴 = 𝐵𝐵𝐶))
8 simpl 486 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴 = 𝐵)
9 eqeltr 38739 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴𝐶)
108, 9jca 519 . 2 ((𝐴 = 𝐵𝐵𝐶) → (𝐴 = 𝐵𝐴𝐶))
117, 10impbii 211 1 ((𝐴 = 𝐵𝐴𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 208  wa 399   = wceq 1560  wcel 2142
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734
This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1800  df-cleq 2754  df-clel 2837
This theorem is referenced by:  eldmressnALTV  38778  raldmqseu  38864  inxpxrn  38917
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