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Theorem eqelb 37565
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 482 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵 = 𝐴)
2 eqeltr 37564 . . . 4 ((𝐵 = 𝐴𝐴𝐶) → 𝐵𝐶)
31, 2jca 511 . . 3 ((𝐵 = 𝐴𝐴𝐶) → (𝐵 = 𝐴𝐵𝐶))
4 eqcom 2738 . . . 4 (𝐵 = 𝐴𝐴 = 𝐵)
54anbi1i 623 . . 3 ((𝐵 = 𝐴𝐴𝐶) ↔ (𝐴 = 𝐵𝐴𝐶))
64anbi1i 623 . . 3 ((𝐵 = 𝐴𝐵𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
73, 5, 63imtr3i 291 . 2 ((𝐴 = 𝐵𝐴𝐶) → (𝐴 = 𝐵𝐵𝐶))
8 simpl 482 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴 = 𝐵)
9 eqeltr 37564 . . 3 ((𝐴 = 𝐵𝐵𝐶) → 𝐴𝐶)
108, 9jca 511 . 2 ((𝐴 = 𝐵𝐵𝐶) → (𝐴 = 𝐵𝐴𝐶))
117, 10impbii 208 1 ((𝐴 = 𝐵𝐴𝐶) ↔ (𝐴 = 𝐵𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wa 395   = wceq 1540  wcel 2105
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702
This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1781  df-cleq 2723  df-clel 2809
This theorem is referenced by:  eldmressnALTV  37604  inxpxrn  37729
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