Metamath Proof Explorer


Theorem srgpcompp

Description: If two elements of a semiring commute, they also commute if the elements are raised to a higher power. (Contributed by AV, 23-Aug-2019)

Ref Expression
Hypotheses srgpcomp.s 𝑆 = ( Base ‘ 𝑅 )
srgpcomp.m × = ( .r𝑅 )
srgpcomp.g 𝐺 = ( mulGrp ‘ 𝑅 )
srgpcomp.e = ( .g𝐺 )
srgpcomp.r ( 𝜑𝑅 ∈ SRing )
srgpcomp.a ( 𝜑𝐴𝑆 )
srgpcomp.b ( 𝜑𝐵𝑆 )
srgpcomp.k ( 𝜑𝐾 ∈ ℕ0 )
srgpcomp.c ( 𝜑 → ( 𝐴 × 𝐵 ) = ( 𝐵 × 𝐴 ) )
srgpcompp.n ( 𝜑𝑁 ∈ ℕ0 )
Assertion srgpcompp ( 𝜑 → ( ( ( 𝑁 𝐴 ) × ( 𝐾 𝐵 ) ) × 𝐴 ) = ( ( ( 𝑁 + 1 ) 𝐴 ) × ( 𝐾 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 srgpcomp.s 𝑆 = ( Base ‘ 𝑅 )
2 srgpcomp.m × = ( .r𝑅 )
3 srgpcomp.g 𝐺 = ( mulGrp ‘ 𝑅 )
4 srgpcomp.e = ( .g𝐺 )
5 srgpcomp.r ( 𝜑𝑅 ∈ SRing )
6 srgpcomp.a ( 𝜑𝐴𝑆 )
7 srgpcomp.b ( 𝜑𝐵𝑆 )
8 srgpcomp.k ( 𝜑𝐾 ∈ ℕ0 )
9 srgpcomp.c ( 𝜑 → ( 𝐴 × 𝐵 ) = ( 𝐵 × 𝐴 ) )
10 srgpcompp.n ( 𝜑𝑁 ∈ ℕ0 )
11 3 srgmgp ( 𝑅 ∈ SRing → 𝐺 ∈ Mnd )
12 5 11 syl ( 𝜑𝐺 ∈ Mnd )
13 3 1 mgpbas 𝑆 = ( Base ‘ 𝐺 )
14 13 4 mulgnn0cl ( ( 𝐺 ∈ Mnd ∧ 𝑁 ∈ ℕ0𝐴𝑆 ) → ( 𝑁 𝐴 ) ∈ 𝑆 )
15 12 10 6 14 syl3anc ( 𝜑 → ( 𝑁 𝐴 ) ∈ 𝑆 )
16 13 4 mulgnn0cl ( ( 𝐺 ∈ Mnd ∧ 𝐾 ∈ ℕ0𝐵𝑆 ) → ( 𝐾 𝐵 ) ∈ 𝑆 )
17 12 8 7 16 syl3anc ( 𝜑 → ( 𝐾 𝐵 ) ∈ 𝑆 )
18 1 2 srgass ( ( 𝑅 ∈ SRing ∧ ( ( 𝑁 𝐴 ) ∈ 𝑆 ∧ ( 𝐾 𝐵 ) ∈ 𝑆𝐴𝑆 ) ) → ( ( ( 𝑁 𝐴 ) × ( 𝐾 𝐵 ) ) × 𝐴 ) = ( ( 𝑁 𝐴 ) × ( ( 𝐾 𝐵 ) × 𝐴 ) ) )
19 5 15 17 6 18 syl13anc ( 𝜑 → ( ( ( 𝑁 𝐴 ) × ( 𝐾 𝐵 ) ) × 𝐴 ) = ( ( 𝑁 𝐴 ) × ( ( 𝐾 𝐵 ) × 𝐴 ) ) )
20 1 2 3 4 5 6 7 8 9 srgpcomp ( 𝜑 → ( ( 𝐾 𝐵 ) × 𝐴 ) = ( 𝐴 × ( 𝐾 𝐵 ) ) )
21 20 oveq2d ( 𝜑 → ( ( 𝑁 𝐴 ) × ( ( 𝐾 𝐵 ) × 𝐴 ) ) = ( ( 𝑁 𝐴 ) × ( 𝐴 × ( 𝐾 𝐵 ) ) ) )
22 1 2 srgass ( ( 𝑅 ∈ SRing ∧ ( ( 𝑁 𝐴 ) ∈ 𝑆𝐴𝑆 ∧ ( 𝐾 𝐵 ) ∈ 𝑆 ) ) → ( ( ( 𝑁 𝐴 ) × 𝐴 ) × ( 𝐾 𝐵 ) ) = ( ( 𝑁 𝐴 ) × ( 𝐴 × ( 𝐾 𝐵 ) ) ) )
23 5 15 6 17 22 syl13anc ( 𝜑 → ( ( ( 𝑁 𝐴 ) × 𝐴 ) × ( 𝐾 𝐵 ) ) = ( ( 𝑁 𝐴 ) × ( 𝐴 × ( 𝐾 𝐵 ) ) ) )
24 21 23 eqtr4d ( 𝜑 → ( ( 𝑁 𝐴 ) × ( ( 𝐾 𝐵 ) × 𝐴 ) ) = ( ( ( 𝑁 𝐴 ) × 𝐴 ) × ( 𝐾 𝐵 ) ) )
25 3 2 mgpplusg × = ( +g𝐺 )
26 13 4 25 mulgnn0p1 ( ( 𝐺 ∈ Mnd ∧ 𝑁 ∈ ℕ0𝐴𝑆 ) → ( ( 𝑁 + 1 ) 𝐴 ) = ( ( 𝑁 𝐴 ) × 𝐴 ) )
27 12 10 6 26 syl3anc ( 𝜑 → ( ( 𝑁 + 1 ) 𝐴 ) = ( ( 𝑁 𝐴 ) × 𝐴 ) )
28 27 eqcomd ( 𝜑 → ( ( 𝑁 𝐴 ) × 𝐴 ) = ( ( 𝑁 + 1 ) 𝐴 ) )
29 28 oveq1d ( 𝜑 → ( ( ( 𝑁 𝐴 ) × 𝐴 ) × ( 𝐾 𝐵 ) ) = ( ( ( 𝑁 + 1 ) 𝐴 ) × ( 𝐾 𝐵 ) ) )
30 19 24 29 3eqtrd ( 𝜑 → ( ( ( 𝑁 𝐴 ) × ( 𝐾 𝐵 ) ) × 𝐴 ) = ( ( ( 𝑁 + 1 ) 𝐴 ) × ( 𝐾 𝐵 ) ) )