Theorem eqbrb 38988

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 488	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) → 𝐵 = 𝐴)
2		eqbrtr 38987	. . . 4 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) → 𝐵𝑅𝐶)
3	1, 2	jca 521	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) → (𝐵 = 𝐴 ∧ 𝐵𝑅𝐶))
4		eqcom 2767	. . . 4 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	anbi1i 636	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐴𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐴𝑅𝐶))
6	4	anbi1i 636	. . 3 ⊢ ((𝐵 = 𝐴 ∧ 𝐵𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))
7	3, 5, 6	3imtr3i 294	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐴𝑅𝐶) → (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))
8		simpl 488	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵𝑅𝐶) → 𝐴 = 𝐵)
9		eqbrtr 38987	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐵𝑅𝐶) → 𝐴𝑅𝐶)
10	8, 9	jca 521	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐵𝑅𝐶) → (𝐴 = 𝐵 ∧ 𝐴𝑅𝐶))
11	7, 10	impbii 212	1 ⊢ ((𝐴 = 𝐵 ∧ 𝐴𝑅𝐶) ↔ (𝐴 = 𝐵 ∧ 𝐵𝑅𝐶))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   class class class wbr 5103

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104

This theorem is used by:  ressn2  39281  trressn  39284