Metamath Proof Explorer


Theorem ex-fac

Description: Example for df-fac . (Contributed by AV, 4-Sep-2021)

Ref Expression
Assertion ex-fac ⊢ 5 ! = 120

Proof

Step Hyp Ref Expression
1 df-5 ⊢ 5 = 4 + 1
2 1 fveq2i ⊢ 5 ! = 4 + 1 !
3 4nn0 ⊢ 4 ∈ ℕ 0
4 facp1 ⊢ 4 ∈ ℕ 0 → 4 + 1 ! = 4 ! ⁢ 4 + 1
5 3 4 ax-mp ⊢ 4 + 1 ! = 4 ! ⁢ 4 + 1
6 2 5 eqtri ⊢ 5 ! = 4 ! ⁢ 4 + 1
7 fac4 ⊢ 4 ! = 24
8 4p1e5 ⊢ 4 + 1 = 5
9 7 8 oveq12i ⊢ 4 ! ⁢ 4 + 1 = 24 ⋅ 5
10 5nn0 ⊢ 5 ∈ ℕ 0
11 2nn0 ⊢ 2 ∈ ℕ 0
12 eqid ⊢ 24 = 24
13 0nn0 ⊢ 0 ∈ ℕ 0
14 1nn0 ⊢ 1 ∈ ℕ 0
15 5cn ⊢ 5 ∈ ℂ
16 2cn ⊢ 2 ∈ ℂ
17 5t2e10 ⊢ 5 ⋅ 2 = 10
18 15 16 17 mulcomli ⊢ 2 ⋅ 5 = 10
19 16 addlidi ⊢ 0 + 2 = 2
20 14 13 11 18 19 decaddi ⊢ 2 ⋅ 5 + 2 = 12
21 4cn ⊢ 4 ∈ ℂ
22 5t4e20 ⊢ 5 ⋅ 4 = 20
23 15 21 22 mulcomli ⊢ 4 ⋅ 5 = 20
24 10 11 3 12 13 11 20 23 decmul1c ⊢ 24 ⋅ 5 = 120
25 9 24 eqtri ⊢ 4 ! ⁢ 4 + 1 = 120
26 6 25 eqtri ⊢ 5 ! = 120