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Theorem eqelb 35935
 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 487 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵 = 𝐴)
2 eqeltr 35934 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵𝐶)
31, 2jca 516 . . 3 ((𝐵 = 𝐴𝐴𝐶) → (𝐵 = 𝐴𝐵𝐶))
4 eqcom 2766 . . . 4 (𝐵 = 𝐴𝐴 = 𝐵)
54anbi1i 627 . . 3 ((𝐵 = 𝐴𝐴𝐶) ↔ (𝐴 = 𝐵𝐴𝐶))
64anbi1i 627 . . 3 ((𝐵 = 𝐴𝐵𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
73, 5, 63imtr3i 295 . 2 ((𝐴 = 𝐵𝐴𝐶) → (𝐴 = 𝐵𝐵𝐶))
8 simpl 487 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴 = 𝐵)
9 eqeltr 35934 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴𝐶)
108, 9jca 516 . 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 1539   ∈ wcel 2112 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 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-ext 2730 This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1783  df-cleq 2751  df-clel 2831 This theorem is referenced by:  inxpxrn  36076
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