Metamath Proof Explorer


Theorem ringm1expp1

Description: Ring exponentiation of minus one: Adding one to the exponent is the same as taking the additive inverse. (Contributed by Thierry Arnoux, 15-Feb-2026)

Ref Expression
Hypotheses ringm1expp1.1 ⊢ 1 ˙ = 1 R
ringm1expp1.2 ⊢ N = inv g ⁡ R
ringm1expp1.3 ⊢ × ˙ = ⋅ mulGrp R
ringm1expp1.4 ⊢ φ → R ∈ Ring
ringm1expp1.5 ⊢ φ → K ∈ ℕ 0
Assertion ringm1expp1 ⊢ φ → K + 1 × ˙ N ⁡ 1 ˙ = N ⁡ K × ˙ N ⁡ 1 ˙

Proof

Step Hyp Ref Expression
1 ringm1expp1.1 ⊢ 1 ˙ = 1 R
2 ringm1expp1.2 ⊢ N = inv g ⁡ R
3 ringm1expp1.3 ⊢ × ˙ = ⋅ mulGrp R
4 ringm1expp1.4 ⊢ φ → R ∈ Ring
5 ringm1expp1.5 ⊢ φ → K ∈ ℕ 0
6 eqid ⊢ mulGrp R = mulGrp R
7 6 ringmgp ⊢ R ∈ Ring → mulGrp R ∈ Mnd
8 4 7 syl ⊢ φ → mulGrp R ∈ Mnd
9 eqid ⊢ Base R = Base R
10 4 ringgrpd ⊢ φ → R ∈ Grp
11 9 1 4 ringidcld ⊢ φ → 1 ˙ ∈ Base R
12 9 2 10 11 grpinvcld ⊢ φ → N ⁡ 1 ˙ ∈ Base R
13 6 9 mgpbas ⊢ Base R = Base mulGrp R
14 eqid ⊢ ⋅ R = ⋅ R
15 6 14 mgpplusg ⊢ ⋅ R = + mulGrp R
16 13 3 15 mulgnn0p1 ⊢ mulGrp R ∈ Mnd ∧ K ∈ ℕ 0 ∧ N ⁡ 1 ˙ ∈ Base R → K + 1 × ˙ N ⁡ 1 ˙ = K × ˙ N ⁡ 1 ˙ ⋅ R N ⁡ 1 ˙
17 8 5 12 16 syl3anc ⊢ φ → K + 1 × ˙ N ⁡ 1 ˙ = K × ˙ N ⁡ 1 ˙ ⋅ R N ⁡ 1 ˙
18 13 3 8 5 12 mulgnn0cld ⊢ φ → K × ˙ N ⁡ 1 ˙ ∈ Base R
19 9 14 1 2 4 18 ringnegr ⊢ φ → K × ˙ N ⁡ 1 ˙ ⋅ R N ⁡ 1 ˙ = N ⁡ K × ˙ N ⁡ 1 ˙
20 17 19 eqtrd ⊢ φ → K + 1 × ˙ N ⁡ 1 ˙ = N ⁡ K × ˙ N ⁡ 1 ˙