Theorem redundeq1 39752

Description: Equivalence of redundancy predicates. (Contributed by Peter Mazsa, 26-Oct-2022.)

Hypothesis

Ref	Expression
redundeq1.1	⊢ 𝐴 = 𝐷

Assertion

Ref	Expression
redundeq1	⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)

Proof of Theorem redundeq1

Step	Hyp	Ref	Expression
1		redundeq1.1	. . . 4 ⊢ 𝐴 = 𝐷
2	1	sseq1i 3937	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ 𝐷 ⊆ 𝐵)
3	1	ineq1i 4112	. . . 4 ⊢ (𝐴 ∩ 𝐶) = (𝐷 ∩ 𝐶)
4	3	eqeq1i 2765	. . 3 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) ↔ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶))
5	2, 4	anbi12i 640	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ (𝐷 ⊆ 𝐵 ∧ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
6		df-redund 39747	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
7		df-redund 39747	. 2 ⊢ (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷 ⊆ 𝐵 ∧ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
8	5, 6, 7	3bitr4i 306	1 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1569   ⊆ wss 3896   ∩ cin 4088   Redund wredund 39243

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-ss 3897  df-in 4090  df-redund 39747

This theorem is used by:  refrelsredund3  39757