Metamath Proof Explorer


Theorem 4bc2eq6

Description: The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017)

Ref Expression
Assertion 4bc2eq6 ( 4 C 2 ) = 6

Proof

Step Hyp Ref Expression
1 0z 0 ∈ ℤ
2 4z 4 ∈ ℤ
3 2z 2 ∈ ℤ
4 1 2 3 3pm3.2i ( 0 ∈ ℤ ∧ 4 ∈ ℤ ∧ 2 ∈ ℤ )
5 0le2 0 ≤ 2
6 2re 2 ∈ ℝ
7 4re 4 ∈ ℝ
8 2lt4 2 < 4
9 6 7 8 ltleii 2 ≤ 4
10 5 9 pm3.2i ( 0 ≤ 2 ∧ 2 ≤ 4 )
11 elfz4 ( ( ( 0 ∈ ℤ ∧ 4 ∈ ℤ ∧ 2 ∈ ℤ ) ∧ ( 0 ≤ 2 ∧ 2 ≤ 4 ) ) → 2 ∈ ( 0 ... 4 ) )
12 4 10 11 mp2an 2 ∈ ( 0 ... 4 )
13 bcval2 ( 2 ∈ ( 0 ... 4 ) → ( 4 C 2 ) = ( ( ! ‘ 4 ) / ( ( ! ‘ ( 4 − 2 ) ) · ( ! ‘ 2 ) ) ) )
14 12 13 ax-mp ( 4 C 2 ) = ( ( ! ‘ 4 ) / ( ( ! ‘ ( 4 − 2 ) ) · ( ! ‘ 2 ) ) )
15 3nn0 3 ∈ ℕ0
16 facp1 ( 3 ∈ ℕ0 → ( ! ‘ ( 3 + 1 ) ) = ( ( ! ‘ 3 ) · ( 3 + 1 ) ) )
17 15 16 ax-mp ( ! ‘ ( 3 + 1 ) ) = ( ( ! ‘ 3 ) · ( 3 + 1 ) )
18 df-4 4 = ( 3 + 1 )
19 18 fveq2i ( ! ‘ 4 ) = ( ! ‘ ( 3 + 1 ) )
20 18 oveq2i ( ( ! ‘ 3 ) · 4 ) = ( ( ! ‘ 3 ) · ( 3 + 1 ) )
21 17 19 20 3eqtr4i ( ! ‘ 4 ) = ( ( ! ‘ 3 ) · 4 )
22 4cn 4 ∈ ℂ
23 2cn 2 ∈ ℂ
24 2p2e4 ( 2 + 2 ) = 4
25 22 23 23 24 subaddrii ( 4 − 2 ) = 2
26 25 fveq2i ( ! ‘ ( 4 − 2 ) ) = ( ! ‘ 2 )
27 fac2 ( ! ‘ 2 ) = 2
28 26 27 eqtri ( ! ‘ ( 4 − 2 ) ) = 2
29 28 27 oveq12i ( ( ! ‘ ( 4 − 2 ) ) · ( ! ‘ 2 ) ) = ( 2 · 2 )
30 2t2e4 ( 2 · 2 ) = 4
31 29 30 eqtri ( ( ! ‘ ( 4 − 2 ) ) · ( ! ‘ 2 ) ) = 4
32 21 31 oveq12i ( ( ! ‘ 4 ) / ( ( ! ‘ ( 4 − 2 ) ) · ( ! ‘ 2 ) ) ) = ( ( ( ! ‘ 3 ) · 4 ) / 4 )
33 faccl ( 3 ∈ ℕ0 → ( ! ‘ 3 ) ∈ ℕ )
34 15 33 ax-mp ( ! ‘ 3 ) ∈ ℕ
35 34 nncni ( ! ‘ 3 ) ∈ ℂ
36 4ne0 4 ≠ 0
37 35 22 36 divcan4i ( ( ( ! ‘ 3 ) · 4 ) / 4 ) = ( ! ‘ 3 )
38 fac3 ( ! ‘ 3 ) = 6
39 37 38 eqtri ( ( ( ! ‘ 3 ) · 4 ) / 4 ) = 6
40 32 39 eqtri ( ( ! ‘ 4 ) / ( ( ! ‘ ( 4 − 2 ) ) · ( ! ‘ 2 ) ) ) = 6
41 14 40 eqtri ( 4 C 2 ) = 6