Metamath Proof Explorer


Theorem fmtno5lem1

Description: Lemma 1 for fmtno5 . (Contributed by AV, 22-Jul-2021)

Ref Expression
Assertion fmtno5lem1 ⊢ 65536 ⋅ 6 = 393216

Proof

Step Hyp Ref Expression
1 6nn0 ⊢ 6 ∈ ℕ 0
2 5nn0 ⊢ 5 ∈ ℕ 0
3 1 2 deccl ⊢ 65 ∈ ℕ 0
4 3 2 deccl ⊢ 655 ∈ ℕ 0
5 3nn0 ⊢ 3 ∈ ℕ 0
6 4 5 deccl ⊢ 6553 ∈ ℕ 0
7 eqid ⊢ 65536 = 65536
8 9nn0 ⊢ 9 ∈ ℕ 0
9 5 8 deccl ⊢ 39 ∈ ℕ 0
10 9 5 deccl ⊢ 393 ∈ ℕ 0
11 1nn0 ⊢ 1 ∈ ℕ 0
12 10 11 deccl ⊢ 3931 ∈ ℕ 0
13 8nn0 ⊢ 8 ∈ ℕ 0
14 eqid ⊢ 6553 = 6553
15 0nn0 ⊢ 0 ∈ ℕ 0
16 0p1e1 ⊢ 0 + 1 = 1
17 eqid ⊢ 655 = 655
18 eqid ⊢ 65 = 65
19 6t6e36 ⊢ 6 ⋅ 6 = 36
20 6p3e9 ⊢ 6 + 3 = 9
21 5 1 5 19 20 decaddi ⊢ 6 ⋅ 6 + 3 = 39
22 6cn ⊢ 6 ∈ ℂ
23 5cn ⊢ 5 ∈ ℂ
24 6t5e30 ⊢ 6 ⋅ 5 = 30
25 22 23 24 mulcomli ⊢ 5 ⋅ 6 = 30
26 1 1 2 18 15 5 21 25 decmul1c ⊢ 65 ⋅ 6 = 390
27 3cn ⊢ 3 ∈ ℂ
28 27 addlidi ⊢ 0 + 3 = 3
29 9 15 5 26 28 decaddi ⊢ 65 ⋅ 6 + 3 = 393
30 1 3 2 17 15 5 29 25 decmul1c ⊢ 655 ⋅ 6 = 3930
31 10 15 16 30 decsuc ⊢ 655 ⋅ 6 + 1 = 3931
32 6t3e18 ⊢ 6 ⋅ 3 = 18
33 22 27 32 mulcomli ⊢ 3 ⋅ 6 = 18
34 1 4 5 14 13 11 31 33 decmul1c ⊢ 6553 ⋅ 6 = 39318
35 1p1e2 ⊢ 1 + 1 = 2
36 eqid ⊢ 3931 = 3931
37 10 11 35 36 decsuc ⊢ 3931 + 1 = 3932
38 8p3e11 ⊢ 8 + 3 = 11
39 12 13 5 34 37 11 38 decaddci ⊢ 6553 ⋅ 6 + 3 = 39321
40 1 6 1 7 1 5 39 19 decmul1c ⊢ 65536 ⋅ 6 = 393216