Metamath Proof Explorer


Theorem faclbnd4lem3

Description: Lemma for faclbnd4 . The N = 0 case. (Contributed by NM, 23-Dec-2005)

Ref Expression
Assertion faclbnd4lem3 ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 ∧ N = 0 → N K ⁢ M N ≤ 2 K 2 ⁢ M M + K ⁢ N !

Proof

Step Hyp Ref Expression
1 elnn0 ⊢ K ∈ ℕ 0 ↔ K ∈ ℕ ∨ K = 0
2 0exp ⊢ K ∈ ℕ → 0 K = 0
3 2 adantl ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ → 0 K = 0
4 nnnn0 ⊢ K ∈ ℕ → K ∈ ℕ 0
5 2nn0 ⊢ 2 ∈ ℕ 0
6 nn0sqcl ⊢ K ∈ ℕ 0 → K 2 ∈ ℕ 0
7 nn0expcl ⊢ 2 ∈ ℕ 0 ∧ K 2 ∈ ℕ 0 → 2 K 2 ∈ ℕ 0
8 5 6 7 sylancr ⊢ K ∈ ℕ 0 → 2 K 2 ∈ ℕ 0
9 8 adantl ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 2 K 2 ∈ ℕ 0
10 nn0addcl ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → M + K ∈ ℕ 0
11 nn0expcl ⊢ M ∈ ℕ 0 ∧ M + K ∈ ℕ 0 → M M + K ∈ ℕ 0
12 10 11 syldan ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → M M + K ∈ ℕ 0
13 9 12 nn0mulcld ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 2 K 2 ⁢ M M + K ∈ ℕ 0
14 4 13 sylan2 ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ → 2 K 2 ⁢ M M + K ∈ ℕ 0
15 14 nn0ge0d ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ → 0 ≤ 2 K 2 ⁢ M M + K
16 3 15 eqbrtrd ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ → 0 K ≤ 2 K 2 ⁢ M M + K
17 1nn ⊢ 1 ∈ ℕ
18 elnn0 ⊢ M ∈ ℕ 0 ↔ M ∈ ℕ ∨ M = 0
19 nnnn0 ⊢ M ∈ ℕ → M ∈ ℕ 0
20 0nn0 ⊢ 0 ∈ ℕ 0
21 nn0addcl ⊢ M ∈ ℕ 0 ∧ 0 ∈ ℕ 0 → M + 0 ∈ ℕ 0
22 19 20 21 sylancl ⊢ M ∈ ℕ → M + 0 ∈ ℕ 0
23 nnexpcl ⊢ M ∈ ℕ ∧ M + 0 ∈ ℕ 0 → M M + 0 ∈ ℕ
24 22 23 mpdan ⊢ M ∈ ℕ → M M + 0 ∈ ℕ
25 id ⊢ M = 0 → M = 0
26 oveq1 ⊢ M = 0 → M + 0 = 0 + 0
27 00id ⊢ 0 + 0 = 0
28 26 27 eqtrdi ⊢ M = 0 → M + 0 = 0
29 25 28 oveq12d ⊢ M = 0 → M M + 0 = 0 0
30 0exp0e1 ⊢ 0 0 = 1
31 29 30 eqtrdi ⊢ M = 0 → M M + 0 = 1
32 31 17 eqeltrdi ⊢ M = 0 → M M + 0 ∈ ℕ
33 24 32 jaoi ⊢ M ∈ ℕ ∨ M = 0 → M M + 0 ∈ ℕ
34 18 33 sylbi ⊢ M ∈ ℕ 0 → M M + 0 ∈ ℕ
35 nnmulcl ⊢ 1 ∈ ℕ ∧ M M + 0 ∈ ℕ → 1 ⁢ M M + 0 ∈ ℕ
36 17 34 35 sylancr ⊢ M ∈ ℕ 0 → 1 ⁢ M M + 0 ∈ ℕ
37 36 nnge1d ⊢ M ∈ ℕ 0 → 1 ≤ 1 ⁢ M M + 0
38 37 adantr ⊢ M ∈ ℕ 0 ∧ K = 0 → 1 ≤ 1 ⁢ M M + 0
39 oveq2 ⊢ K = 0 → 0 K = 0 0
40 39 30 eqtrdi ⊢ K = 0 → 0 K = 1
41 sq0i ⊢ K = 0 → K 2 = 0
42 41 oveq2d ⊢ K = 0 → 2 K 2 = 2 0
43 2cn ⊢ 2 ∈ ℂ
44 exp0 ⊢ 2 ∈ ℂ → 2 0 = 1
45 43 44 ax-mp ⊢ 2 0 = 1
46 42 45 eqtrdi ⊢ K = 0 → 2 K 2 = 1
47 oveq2 ⊢ K = 0 → M + K = M + 0
48 47 oveq2d ⊢ K = 0 → M M + K = M M + 0
49 46 48 oveq12d ⊢ K = 0 → 2 K 2 ⁢ M M + K = 1 ⁢ M M + 0
50 40 49 breq12d ⊢ K = 0 → 0 K ≤ 2 K 2 ⁢ M M + K ↔ 1 ≤ 1 ⁢ M M + 0
51 50 adantl ⊢ M ∈ ℕ 0 ∧ K = 0 → 0 K ≤ 2 K 2 ⁢ M M + K ↔ 1 ≤ 1 ⁢ M M + 0
52 38 51 mpbird ⊢ M ∈ ℕ 0 ∧ K = 0 → 0 K ≤ 2 K 2 ⁢ M M + K
53 16 52 jaodan ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ ∨ K = 0 → 0 K ≤ 2 K 2 ⁢ M M + K
54 1 53 sylan2b ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 0 K ≤ 2 K 2 ⁢ M M + K
55 nn0cn ⊢ M ∈ ℕ 0 → M ∈ ℂ
56 55 exp0d ⊢ M ∈ ℕ 0 → M 0 = 1
57 56 oveq2d ⊢ M ∈ ℕ 0 → 0 K ⁢ M 0 = 0 K ⋅ 1
58 nn0expcl ⊢ 0 ∈ ℕ 0 ∧ K ∈ ℕ 0 → 0 K ∈ ℕ 0
59 20 58 mpan ⊢ K ∈ ℕ 0 → 0 K ∈ ℕ 0
60 59 nn0cnd ⊢ K ∈ ℕ 0 → 0 K ∈ ℂ
61 60 mulridd ⊢ K ∈ ℕ 0 → 0 K ⋅ 1 = 0 K
62 57 61 sylan9eq ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 0 K ⁢ M 0 = 0 K
63 13 nn0cnd ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 2 K 2 ⁢ M M + K ∈ ℂ
64 63 mulridd ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 2 K 2 ⁢ M M + K ⋅ 1 = 2 K 2 ⁢ M M + K
65 54 62 64 3brtr4d ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 → 0 K ⁢ M 0 ≤ 2 K 2 ⁢ M M + K ⋅ 1
66 65 adantr ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 ∧ N = 0 → 0 K ⁢ M 0 ≤ 2 K 2 ⁢ M M + K ⋅ 1
67 oveq1 ⊢ N = 0 → N K = 0 K
68 oveq2 ⊢ N = 0 → M N = M 0
69 67 68 oveq12d ⊢ N = 0 → N K ⁢ M N = 0 K ⁢ M 0
70 fveq2 ⊢ N = 0 → N ! = 0 !
71 fac0 ⊢ 0 ! = 1
72 70 71 eqtrdi ⊢ N = 0 → N ! = 1
73 72 oveq2d ⊢ N = 0 → 2 K 2 ⁢ M M + K ⁢ N ! = 2 K 2 ⁢ M M + K ⋅ 1
74 69 73 breq12d ⊢ N = 0 → N K ⁢ M N ≤ 2 K 2 ⁢ M M + K ⁢ N ! ↔ 0 K ⁢ M 0 ≤ 2 K 2 ⁢ M M + K ⋅ 1
75 74 adantl ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 ∧ N = 0 → N K ⁢ M N ≤ 2 K 2 ⁢ M M + K ⁢ N ! ↔ 0 K ⁢ M 0 ≤ 2 K 2 ⁢ M M + K ⋅ 1
76 66 75 mpbird ⊢ M ∈ ℕ 0 ∧ K ∈ ℕ 0 ∧ N = 0 → N K ⁢ M N ≤ 2 K 2 ⁢ M M + K ⁢ N !