Theorem redundeq1 38666

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 3958	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ 𝐷 ⊆ 𝐵)
3	1	ineq1i 4161	. . . 4 ⊢ (𝐴 ∩ 𝐶) = (𝐷 ∩ 𝐶)
4	3	eqeq1i 2736	. . 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 38661	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
7		df-redund 38661	. 2 ⊢ (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷 ⊆ 𝐵 ∧ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
8	5, 6, 7	3bitr4i 303	1 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1541   ∩ cin 3896   ⊆ wss 3897   Redund wredund 38236

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-in 3904  df-ss 3914  df-redund 38661

This theorem is referenced by:  refrelsredund3  38671