Theorem redundss3 35897

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

Hypothesis

Ref	Expression
redundss3.1	⊢ 𝐷 ⊆ 𝐶

Assertion

Ref	Expression
redundss3	⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ → 𝐴 Redund ⟨𝐵, 𝐷⟩)

Proof of Theorem redundss3

Step	Hyp	Ref	Expression
1		ineq1 4174	. . . 4 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → ((𝐴 ∩ 𝐶) ∩ 𝐷) = ((𝐵 ∩ 𝐶) ∩ 𝐷))
2		redundss3.1	. . . . . . . 8 ⊢ 𝐷 ⊆ 𝐶
3		dfss 3946	. . . . . . . 8 ⊢ (𝐷 ⊆ 𝐶 ↔ 𝐷 = (𝐷 ∩ 𝐶))
4	2, 3	mpbi 232	. . . . . . 7 ⊢ 𝐷 = (𝐷 ∩ 𝐶)
5		incom 4171	. . . . . . 7 ⊢ (𝐷 ∩ 𝐶) = (𝐶 ∩ 𝐷)
6	4, 5	eqtri 2843	. . . . . 6 ⊢ 𝐷 = (𝐶 ∩ 𝐷)
7	6	ineq2i 4179	. . . . 5 ⊢ (𝐴 ∩ 𝐷) = (𝐴 ∩ (𝐶 ∩ 𝐷))
8		inass 4189	. . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∩ 𝐷) = (𝐴 ∩ (𝐶 ∩ 𝐷))
9	7, 8	eqtr4i 2846	. . . 4 ⊢ (𝐴 ∩ 𝐷) = ((𝐴 ∩ 𝐶) ∩ 𝐷)
10	6	ineq2i 4179	. . . . 5 ⊢ (𝐵 ∩ 𝐷) = (𝐵 ∩ (𝐶 ∩ 𝐷))
11		inass 4189	. . . . 5 ⊢ ((𝐵 ∩ 𝐶) ∩ 𝐷) = (𝐵 ∩ (𝐶 ∩ 𝐷))
12	10, 11	eqtr4i 2846	. . . 4 ⊢ (𝐵 ∩ 𝐷) = ((𝐵 ∩ 𝐶) ∩ 𝐷)
13	1, 9, 12	3eqtr4g 2880	. . 3 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝐴 ∩ 𝐷) = (𝐵 ∩ 𝐷))
14	13	anim2i 618	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐷) = (𝐵 ∩ 𝐷)))
15		df-redund 35893	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
16		df-redund 35893	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐷⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐷) = (𝐵 ∩ 𝐷)))
17	14, 15, 16	3imtr4i 294	1 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ → 𝐴 Redund ⟨𝐵, 𝐷⟩)

Syntax hints:   → wi 4   ∧ wa 398   = wceq 1536   ∩ cin 3928   ⊆ wss 3929   Redund wredund 35508

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-nfc 2962  df-rab 3146  df-v 3493  df-in 3936  df-ss 3945  df-redund 35893

This theorem is referenced by:  refrelsredund2  35902