Users' Mathboxes Mathbox for Peter Mazsa < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  eqbrb Structured version   Visualization version   GIF version

Theorem eqbrb 38490
Description: Substitution of equal classes in a binary relation. (Contributed by Peter Mazsa, 14-Jun-2024.)
Assertion
Ref Expression
eqbrb ((𝐴 = 𝐵𝐴𝑅𝐶) ↔ (𝐴 = 𝐵𝐵𝑅𝐶))

Proof of Theorem eqbrb
StepHypRef Expression
1 simpl 482 . . . 4 ((𝐵 = 𝐴𝐴𝑅𝐶) → 𝐵 = 𝐴)
2 eqbrtr 38489 . . . 4 ((𝐵 = 𝐴𝐴𝑅𝐶) → 𝐵𝑅𝐶)
31, 2jca 511 . . 3 ((𝐵 = 𝐴𝐴𝑅𝐶) → (𝐵 = 𝐴𝐵𝑅𝐶))
4 eqcom 2744 . . . 4 (𝐵 = 𝐴𝐴 = 𝐵)
54anbi1i 625 . . 3 ((𝐵 = 𝐴𝐴𝑅𝐶) ↔ (𝐴 = 𝐵𝐴𝑅𝐶))
64anbi1i 625 . . 3 ((𝐵 = 𝐴𝐵𝑅𝐶) ↔ (𝐴 = 𝐵𝐵𝑅𝐶))
73, 5, 63imtr3i 291 . 2 ((𝐴 = 𝐵𝐴𝑅𝐶) → (𝐴 = 𝐵𝐵𝑅𝐶))
8 simpl 482 . . 3 ((𝐴 = 𝐵𝐵𝑅𝐶) → 𝐴 = 𝐵)
9 eqbrtr 38489 . . 3 ((𝐴 = 𝐵𝐵𝑅𝐶) → 𝐴𝑅𝐶)
108, 9jca 511 . 2 ((𝐴 = 𝐵𝐵𝑅𝐶) → (𝐴 = 𝐵𝐴𝑅𝐶))
117, 10impbii 209 1 ((𝐴 = 𝐵𝐴𝑅𝐶) ↔ (𝐴 = 𝐵𝐵𝑅𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395   = wceq 1542   class class class wbr 5100
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101
This theorem is referenced by:  ressn2  38783  trressn  38786
  Copyright terms: Public domain W3C validator