Metamath Proof Explorer


Theorem ssdifidl

Description: Let R be a ring, and let I be an ideal of R disjoint with a set S . Then there exists an ideal i , maximal among the set P of ideals containing I and disjoint with S . (Contributed by Thierry Arnoux, 3-Jun-2025)

Ref Expression
Hypotheses ssdifidl.1 ⊢ B = Base R
ssdifidl.2 ⊢ φ → R ∈ Ring
ssdifidl.3 ⊢ φ → I ∈ LIdeal ⁡ R
ssdifidl.4 ⊢ φ → S ⊆ B
ssdifidl.5 ⊢ φ → S ∩ I = ∅
ssdifidl.6 ⊢ P = p ∈ LIdeal ⁡ R | S ∩ p = ∅ ∧ I ⊆ p
Assertion ssdifidl ⊢ φ → ∃ i ∈ P ∀ j ∈ P ¬ i ⊂ j

Proof

Step Hyp Ref Expression
1 ssdifidl.1 ⊢ B = Base R
2 ssdifidl.2 ⊢ φ → R ∈ Ring
3 ssdifidl.3 ⊢ φ → I ∈ LIdeal ⁡ R
4 ssdifidl.4 ⊢ φ → S ⊆ B
5 ssdifidl.5 ⊢ φ → S ∩ I = ∅
6 ssdifidl.6 ⊢ P = p ∈ LIdeal ⁡ R | S ∩ p = ∅ ∧ I ⊆ p
7 ineq2 ⊢ p = I → S ∩ p = S ∩ I
8 7 eqeq1d ⊢ p = I → S ∩ p = ∅ ↔ S ∩ I = ∅
9 sseq2 ⊢ p = I → I ⊆ p ↔ I ⊆ I
10 8 9 anbi12d ⊢ p = I → S ∩ p = ∅ ∧ I ⊆ p ↔ S ∩ I = ∅ ∧ I ⊆ I
11 ssidd ⊢ φ → I ⊆ I
12 5 11 jca ⊢ φ → S ∩ I = ∅ ∧ I ⊆ I
13 10 3 12 elrabd ⊢ φ → I ∈ p ∈ LIdeal ⁡ R | S ∩ p = ∅ ∧ I ⊆ p
14 13 6 eleqtrrdi ⊢ φ → I ∈ P
15 14 ne0d ⊢ φ → P ≠ ∅
16 2 adantr ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → R ∈ Ring
17 3 adantr ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → I ∈ LIdeal ⁡ R
18 4 adantr ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → S ⊆ B
19 5 adantr ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → S ∩ I = ∅
20 simpr1 ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → z ⊆ P
21 simpr2 ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → z ≠ ∅
22 simpr3 ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → [⊂] Or z
23 1 16 17 18 19 6 20 21 22 ssdifidllem ⊢ φ ∧ z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → ⋃ z ∈ P
24 23 ex ⊢ φ → z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → ⋃ z ∈ P
25 24 alrimiv ⊢ φ → ∀ z z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → ⋃ z ∈ P
26 fvex ⊢ LIdeal ⁡ R ∈ V
27 6 26 rabex2 ⊢ P ∈ V
28 27 zornn0 ⊢ P ≠ ∅ ∧ ∀ z z ⊆ P ∧ z ≠ ∅ ∧ [⊂] Or z → ⋃ z ∈ P → ∃ i ∈ P ∀ j ∈ P ¬ i ⊂ j
29 15 25 28 syl2anc ⊢ φ → ∃ i ∈ P ∀ j ∈ P ¬ i ⊂ j