Theorem eqbrb 38888

Description: Substitution of equal classes in a binary relation. (Contributed by Peter Mazsa, 14-Jun-2024.)

Assertion

Ref	Expression
eqbrb	⊢ ((𝐴 = 𝐵 ∧ 𝐴𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))

Proof of Theorem eqbrb

Step	Hyp	Ref	Expression
1		simpl 487	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) → 𝐵 = 𝐴)
2		eqbrtr 38887	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) → 𝐵𝑅𝐶)
3	1, 2	jca 520	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) → (𝐵 = 𝐴 ∧ 𝐵𝑅𝐶))
4		eqcom 2770	. . . 4 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	anbi1i 635	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐴𝑅𝐶))
6	4	anbi1i 635	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐵𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))
7	3, 5, 6	3imtr3i 294	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴𝑅𝐶) → (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))
8		simpl 487	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵𝑅𝐶) → 𝐴 = 𝐵)
9		eqbrtr 38887	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵𝑅𝐶) → 𝐴𝑅𝐶)
10	8, 9	jca 520	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵𝑅𝐶) → (𝐴 = 𝐵 ∧ 𝐴𝑅𝐶))
11	7, 10	impbii 212	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))

Syntax hints:   ↔ wb 209   ∧ wa 400   = wceq 1570   class class class wbr 5109

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-br 5110

This theorem is referenced by:  ressn2  39181  trressn  39184