Theorem coineq0 38201

Description: A composition (𝐴 ∘ 𝐵) is disjoint from a class 𝐶 if and only if 𝐵 is disjoint from (^◡𝐴 ∘ 𝐶). This is a form of the Schroeder rule that does not use complements. See also coeq0 6274. (Contributed by BJ, 8-Nov-2026.)

Assertion

Ref	Expression
coineq0	⊢ (((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅ ↔ (𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ∅)

Proof of Theorem coineq0

Step	Hyp	Ref	Expression
1		incom 4109	. . 3 ⊢ (𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ((◡𝐴 ∘ 𝐶) ∩ 𝐵)
2		cocnvcnv1 6276	. . . . . . 7 ⊢ (◡◡𝐴 ∘ 𝐵) = (𝐴 ∘ 𝐵)
3	2	ineq2i 4117	. . . . . 6 ⊢ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵)) = (𝐶 ∩ (𝐴 ∘ 𝐵))
4		incom 4109	. . . . . 6 ⊢ (𝐶 ∩ (𝐴 ∘ 𝐵)) = ((𝐴 ∘ 𝐵) ∩ 𝐶)
5	3, 4	eqtri 2786	. . . . 5 ⊢ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵)) = ((𝐴 ∘ 𝐵) ∩ 𝐶)
6		id 23	. . . . 5 ⊢ (((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅ → ((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅)
7	5, 6	eqtrid 2810	. . . 4 ⊢ (((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅ → (𝐶 ∩ (◡◡𝐴 ∘ 𝐵)) = ∅)
8		coeq2 5845	. . . . . 6 ⊢ ((𝐶 ∩ (◡◡𝐴 ∘ 𝐵)) = ∅ → ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵))) = ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ ∅))
9		co02 6281	. . . . . 6 ⊢ ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ ∅) = ∅
10	8, 9	eqtrdi 2814	. . . . 5 ⊢ ((𝐶 ∩ (◡◡𝐴 ∘ 𝐵)) = ∅ → ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵))) = ∅)
11		comodular 38200	. . . . . 6 ⊢ ((◡𝐴 ∘ 𝐶) ∩ 𝐵) ⊆ ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵)))
12		sseq0 4357	. . . . . 6 ⊢ ((((◡𝐴 ∘ 𝐶) ∩ 𝐵) ⊆ ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵))) ∧ ((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵))) = ∅) → ((◡𝐴 ∘ 𝐶) ∩ 𝐵) = ∅)
13	11, 12	mpan 703	. . . . 5 ⊢ (((◡𝐴 ∩ (𝐵 ∘ ◡𝐶)) ∘ (𝐶 ∩ (◡◡𝐴 ∘ 𝐵))) = ∅ → ((◡𝐴 ∘ 𝐶) ∩ 𝐵) = ∅)
14	10, 13	syl 18	. . . 4 ⊢ ((𝐶 ∩ (◡◡𝐴 ∘ 𝐵)) = ∅ → ((◡𝐴 ∘ 𝐶) ∩ 𝐵) = ∅)
15	7, 14	syl 18	. . 3 ⊢ (((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅ → ((◡𝐴 ∘ 𝐶) ∩ 𝐵) = ∅)
16	1, 15	eqtrid 2810	. 2 ⊢ (((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅ → (𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ∅)
17		coeq2 5845	. . . 4 ⊢ ((𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ∅ → ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ (𝐵 ∩ (◡𝐴 ∘ 𝐶))) = ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ ∅))
18		co02 6281	. . . 4 ⊢ ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ ∅) = ∅
19	17, 18	eqtrdi 2814	. . 3 ⊢ ((𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ∅ → ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ (𝐵 ∩ (◡𝐴 ∘ 𝐶))) = ∅)
20		comodular 38200	. . . 4 ⊢ ((𝐴 ∘ 𝐵) ∩ 𝐶) ⊆ ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ (𝐵 ∩ (◡𝐴 ∘ 𝐶)))
21		sseq0 4357	. . . 4 ⊢ ((((𝐴 ∘ 𝐵) ∩ 𝐶) ⊆ ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ (𝐵 ∩ (◡𝐴 ∘ 𝐶))) ∧ ((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ (𝐵 ∩ (◡𝐴 ∘ 𝐶))) = ∅) → ((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅)
22	20, 21	mpan 703	. . 3 ⊢ (((𝐴 ∩ (𝐶 ∘ ◡𝐵)) ∘ (𝐵 ∩ (◡𝐴 ∘ 𝐶))) = ∅ → ((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅)
23	19, 22	syl 18	. 2 ⊢ ((𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ∅ → ((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅)
24	16, 23	impbii 212	1 ⊢ (((𝐴 ∘ 𝐵) ∩ 𝐶) = ∅ ↔ (𝐵 ∩ (◡𝐴 ∘ 𝐶)) = ∅)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ⊆ wss 3900   ∩ cin 4092  ∅c0 4281  ^◡ccnv 5655   ∘ ccom 5660

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5253  ax-pr 5396

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-xp 5662  df-rel 5663  df-cnv 5664  df-co 5665  df-dm 5666  df-rn 5667  df-res 5668

This theorem is used by: (None)