Theorem redundss3 39401

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 4169	. . . 4 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → ((𝐴 ∩ 𝐶) ∩ 𝐷) = ((𝐵 ∩ 𝐶) ∩ 𝐷))
2		redundss3.1	. . . . . . . 8 ⊢ 𝐷 ⊆ 𝐶
3		dfss 3927	. . . . . . . 8 ⊢ (𝐷 ⊆ 𝐶 ↔ 𝐷 = (𝐷 ∩ 𝐶))
4	2, 3	mpbi 233	. . . . . . 7 ⊢ 𝐷 = (𝐷 ∩ 𝐶)
5		incom 4165	. . . . . . 7 ⊢ (𝐷 ∩ 𝐶) = (𝐶 ∩ 𝐷)
6	4, 5	eqtri 2789	. . . . . 6 ⊢ 𝐷 = (𝐶 ∩ 𝐷)
7	6	ineq2i 4173	. . . . 5 ⊢ (𝐴 ∩ 𝐷) = (𝐴 ∩ (𝐶 ∩ 𝐷))
8		inass 4183	. . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∩ 𝐷) = (𝐴 ∩ (𝐶 ∩ 𝐷))
9	7, 8	eqtr4i 2792	. . . 4 ⊢ (𝐴 ∩ 𝐷) = ((𝐴 ∩ 𝐶) ∩ 𝐷)
10	6	ineq2i 4173	. . . . 5 ⊢ (𝐵 ∩ 𝐷) = (𝐵 ∩ (𝐶 ∩ 𝐷))
11		inass 4183	. . . . 5 ⊢ ((𝐵 ∩ 𝐶) ∩ 𝐷) = (𝐵 ∩ (𝐶 ∩ 𝐷))
12	10, 11	eqtr4i 2792	. . . 4 ⊢ (𝐵 ∩ 𝐷) = ((𝐵 ∩ 𝐶) ∩ 𝐷)
13	1, 9, 12	3eqtr4g 2826	. . 3 ⊢ ((𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶) → (𝐴 ∩ 𝐷) = (𝐵 ∩ 𝐷))
14	13	anim2i 629	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)) → (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐷) = (𝐵 ∩ 𝐷)))
15		df-redund 39397	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐶) = (𝐵 ∩ 𝐶)))
16		df-redund 39397	. 2 ⊢ (𝐴 Redund ⟨𝐵, 𝐷⟩ ↔ (𝐴 ⊆ 𝐵 ∧ (𝐴 ∩ 𝐷) = (𝐵 ∩ 𝐷)))
17	14, 15, 16	3imtr4i 295	1 ⊢ (𝐴 Redund ⟨𝐵, 𝐶⟩ → 𝐴 Redund ⟨𝐵, 𝐷⟩)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∩ cin 3907   ⊆ wss 3908   Redund wredund 38893

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-rab 3420  df-v 3460  df-in 3915  df-ss 3925  df-redund 39397

This theorem is used by:  refrelsredund2  39406