Metamath Proof Explorer


Theorem srgbinomlem2

Description: Lemma 2 for srgbinomlem . (Contributed by AV, 23-Aug-2019)

Ref Expression
Hypotheses srgbinom.s 𝑆 = ( Base ‘ 𝑅 )
srgbinom.m × = ( .r𝑅 )
srgbinom.t · = ( .g𝑅 )
srgbinom.a + = ( +g𝑅 )
srgbinom.g 𝐺 = ( mulGrp ‘ 𝑅 )
srgbinom.e = ( .g𝐺 )
srgbinomlem.r ( 𝜑𝑅 ∈ SRing )
srgbinomlem.a ( 𝜑𝐴𝑆 )
srgbinomlem.b ( 𝜑𝐵𝑆 )
srgbinomlem.c ( 𝜑 → ( 𝐴 × 𝐵 ) = ( 𝐵 × 𝐴 ) )
srgbinomlem.n ( 𝜑𝑁 ∈ ℕ0 )
Assertion srgbinomlem2 ( ( 𝜑 ∧ ( 𝐶 ∈ ℕ0𝐷 ∈ ℕ0𝐸 ∈ ℕ0 ) ) → ( 𝐶 · ( ( 𝐷 𝐴 ) × ( 𝐸 𝐵 ) ) ) ∈ 𝑆 )

Proof

Step Hyp Ref Expression
1 srgbinom.s 𝑆 = ( Base ‘ 𝑅 )
2 srgbinom.m × = ( .r𝑅 )
3 srgbinom.t · = ( .g𝑅 )
4 srgbinom.a + = ( +g𝑅 )
5 srgbinom.g 𝐺 = ( mulGrp ‘ 𝑅 )
6 srgbinom.e = ( .g𝐺 )
7 srgbinomlem.r ( 𝜑𝑅 ∈ SRing )
8 srgbinomlem.a ( 𝜑𝐴𝑆 )
9 srgbinomlem.b ( 𝜑𝐵𝑆 )
10 srgbinomlem.c ( 𝜑 → ( 𝐴 × 𝐵 ) = ( 𝐵 × 𝐴 ) )
11 srgbinomlem.n ( 𝜑𝑁 ∈ ℕ0 )
12 srgmnd ( 𝑅 ∈ SRing → 𝑅 ∈ Mnd )
13 7 12 syl ( 𝜑𝑅 ∈ Mnd )
14 13 adantr ( ( 𝜑 ∧ ( 𝐶 ∈ ℕ0𝐷 ∈ ℕ0𝐸 ∈ ℕ0 ) ) → 𝑅 ∈ Mnd )
15 simpr1 ( ( 𝜑 ∧ ( 𝐶 ∈ ℕ0𝐷 ∈ ℕ0𝐸 ∈ ℕ0 ) ) → 𝐶 ∈ ℕ0 )
16 1 2 3 4 5 6 7 8 9 10 11 srgbinomlem1 ( ( 𝜑 ∧ ( 𝐷 ∈ ℕ0𝐸 ∈ ℕ0 ) ) → ( ( 𝐷 𝐴 ) × ( 𝐸 𝐵 ) ) ∈ 𝑆 )
17 16 3adantr1 ( ( 𝜑 ∧ ( 𝐶 ∈ ℕ0𝐷 ∈ ℕ0𝐸 ∈ ℕ0 ) ) → ( ( 𝐷 𝐴 ) × ( 𝐸 𝐵 ) ) ∈ 𝑆 )
18 1 3 mulgnn0cl ( ( 𝑅 ∈ Mnd ∧ 𝐶 ∈ ℕ0 ∧ ( ( 𝐷 𝐴 ) × ( 𝐸 𝐵 ) ) ∈ 𝑆 ) → ( 𝐶 · ( ( 𝐷 𝐴 ) × ( 𝐸 𝐵 ) ) ) ∈ 𝑆 )
19 14 15 17 18 syl3anc ( ( 𝜑 ∧ ( 𝐶 ∈ ℕ0𝐷 ∈ ℕ0𝐸 ∈ ℕ0 ) ) → ( 𝐶 · ( ( 𝐷 𝐴 ) × ( 𝐸 𝐵 ) ) ) ∈ 𝑆 )