Metamath Proof Explorer


Theorem 0ringsubrg

Description: A subring of a zero ring is a zero ring. (Contributed by Thierry Arnoux, 5-Feb-2025)

Ref Expression
Hypotheses 0ringsubrg.1 ⊢ B = Base R
0ringsubrg.2 ⊢ φ → R ∈ Ring
0ringsubrg.3 ⊢ φ → B = 1
0ringsubrg.4 ⊢ φ → S ∈ SubRing ⁡ R
Assertion 0ringsubrg ⊢ φ → S = 1

Proof

Step Hyp Ref Expression
1 0ringsubrg.1 ⊢ B = Base R
2 0ringsubrg.2 ⊢ φ → R ∈ Ring
3 0ringsubrg.3 ⊢ φ → B = 1
4 0ringsubrg.4 ⊢ φ → S ∈ SubRing ⁡ R
5 1 subrgss ⊢ S ∈ SubRing ⁡ R → S ⊆ B
6 4 5 syl ⊢ φ → S ⊆ B
7 eqid ⊢ 0 R = 0 R
8 1 7 0ring ⊢ R ∈ Ring ∧ B = 1 → B = 0 R
9 2 3 8 syl2anc ⊢ φ → B = 0 R
10 6 9 sseqtrd ⊢ φ → S ⊆ 0 R
11 sssn ⊢ S ⊆ 0 R ↔ S = ∅ ∨ S = 0 R
12 10 11 sylib ⊢ φ → S = ∅ ∨ S = 0 R
13 eqid ⊢ 1 R = 1 R
14 13 subrg1cl ⊢ S ∈ SubRing ⁡ R → 1 R ∈ S
15 4 14 syl ⊢ φ → 1 R ∈ S
16 n0i ⊢ 1 R ∈ S → ¬ S = ∅
17 15 16 syl ⊢ φ → ¬ S = ∅
18 12 17 orcnd ⊢ φ → S = 0 R
19 18 fveq2d ⊢ φ → S = 0 R
20 fvex ⊢ 0 R ∈ V
21 hashsng ⊢ 0 R ∈ V → 0 R = 1
22 20 21 ax-mp ⊢ 0 R = 1
23 19 22 eqtrdi ⊢ φ → S = 1