Metamath Proof Explorer


Theorem bcn2m1

Description: Compute the binomial coefficient " N choose 2 " from " ( N - 1 ) choose 2 ": (N-1) + ( (N-1) 2 ) = ( N 2 ). (Contributed by Alexander van der Vekens, 7-Jan-2018)

Ref Expression
Assertion bcn2m1 ⊢ N ∈ ℕ → N - 1 + ( N − 1 2 ) = ( N 2 )

Proof

Step Hyp Ref Expression
1 nnm1nn0 ⊢ N ∈ ℕ → N − 1 ∈ ℕ 0
2 1 nn0cnd ⊢ N ∈ ℕ → N − 1 ∈ ℂ
3 2z ⊢ 2 ∈ ℤ
4 bccl ⊢ N − 1 ∈ ℕ 0 ∧ 2 ∈ ℤ → ( N − 1 2 ) ∈ ℕ 0
5 1 3 4 sylancl ⊢ N ∈ ℕ → ( N − 1 2 ) ∈ ℕ 0
6 5 nn0cnd ⊢ N ∈ ℕ → ( N − 1 2 ) ∈ ℂ
7 2 6 addcomd ⊢ N ∈ ℕ → N - 1 + ( N − 1 2 ) = ( N − 1 2 ) + N - 1
8 bcn1 ⊢ N − 1 ∈ ℕ 0 → ( N − 1 1 ) = N − 1
9 8 eqcomd ⊢ N − 1 ∈ ℕ 0 → N − 1 = ( N − 1 1 )
10 1 9 syl ⊢ N ∈ ℕ → N − 1 = ( N − 1 1 )
11 1e2m1 ⊢ 1 = 2 − 1
12 11 a1i ⊢ N ∈ ℕ → 1 = 2 − 1
13 12 oveq2d ⊢ N ∈ ℕ → ( N − 1 1 ) = ( N − 1 2 − 1 )
14 10 13 eqtrd ⊢ N ∈ ℕ → N − 1 = ( N − 1 2 − 1 )
15 14 oveq2d ⊢ N ∈ ℕ → ( N − 1 2 ) + N - 1 = ( N − 1 2 ) + ( N − 1 2 − 1 )
16 bcpasc ⊢ N − 1 ∈ ℕ 0 ∧ 2 ∈ ℤ → ( N − 1 2 ) + ( N − 1 2 − 1 ) = ( N - 1 + 1 2 )
17 1 3 16 sylancl ⊢ N ∈ ℕ → ( N − 1 2 ) + ( N − 1 2 − 1 ) = ( N - 1 + 1 2 )
18 nncn ⊢ N ∈ ℕ → N ∈ ℂ
19 1cnd ⊢ N ∈ ℕ → 1 ∈ ℂ
20 18 19 npcand ⊢ N ∈ ℕ → N - 1 + 1 = N
21 20 oveq1d ⊢ N ∈ ℕ → ( N - 1 + 1 2 ) = ( N 2 )
22 17 21 eqtrd ⊢ N ∈ ℕ → ( N − 1 2 ) + ( N − 1 2 − 1 ) = ( N 2 )
23 7 15 22 3eqtrd ⊢ N ∈ ℕ → N - 1 + ( N − 1 2 ) = ( N 2 )