Theorem redundeq1 36385

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 3905	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ 𝐷 ⊆ 𝐵)
3	1	ineq1i 4099	. . . 4 ⊢ (𝐴 ∩ 𝐶) = (𝐷 ∩ 𝐶)
4	3	eqeq1i 2743	. . 3 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) ↔ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶))
5	2, 4	anbi12i 630	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) ↔ (𝐷 ⊆ 𝐵 ∧ (𝐷 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
6		df-redund 36380	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
7		df-redund 36380	. 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⟩)

Syntax hints:   ↔ wb 209   ∧ wa 399   = wceq 1542   ∩ cin 3842   ⊆ wss 3843   Redund wredund 35997

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-ex 1787  df-sb 2075  df-clab 2717  df-cleq 2730  df-clel 2811  df-rab 3062  df-v 3400  df-in 3850  df-ss 3860  df-redund 36380

This theorem is referenced by:  refrelsredund3  36390