Theorem redundeq1 35897

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 3988	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ 𝐷 ⊆ 𝐵)
3	1	ineq1i 4178	. . . 4 ⊢ (𝐴 ∩ 𝐶) = (𝐷 ∩ 𝐶)
4	3	eqeq1i 2825	. . 3 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) ↔ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶))
5	2, 4	anbi12i 628	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ (𝐷 ⊆ 𝐵 ∧ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
6		df-redund 35892	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
7		df-redund 35892	. 2 ⊢ (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷 ⊆ 𝐵 ∧ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
8	5, 6, 7	3bitr4i 305	1 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)

Syntax hints:   ↔ wb 208   ∧ wa 398   = wceq 1536   ∩ cin 3928   ⊆ wss 3929   Redund wredund 35507

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2792

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-clab 2799  df-cleq 2813  df-clel 2892  df-rab 3146  df-in 3936  df-ss 3945  df-redund 35892

This theorem is referenced by:  refrelsredund3  35902