Metamath Proof Explorer


Theorem bcnn

Description: N choose N is 1. Remark in Gleason p. 296. (Contributed by NM, 17-Jun-2005) (Revised by Mario Carneiro, 8-Nov-2013)

Ref Expression
Assertion bcnn ( 𝑁 ∈ ℕ0 → ( 𝑁 C 𝑁 ) = 1 )

Proof

Step Hyp Ref Expression
1 0z ⊢ 0 ∈ ℤ
2 bccmpl ⊢ ( ( 𝑁 ∈ ℕ0 ∧ 0 ∈ ℤ ) → ( 𝑁 C 0 ) = ( 𝑁 C ( 𝑁 − 0 ) ) )
3 1 2 mpan2 ⊢ ( 𝑁 ∈ ℕ0 → ( 𝑁 C 0 ) = ( 𝑁 C ( 𝑁 − 0 ) ) )
4 bcn0 ⊢ ( 𝑁 ∈ ℕ0 → ( 𝑁 C 0 ) = 1 )
5 nn0cn ⊢ ( 𝑁 ∈ ℕ0 → 𝑁 ∈ ℂ )
6 5 subid1d ⊢ ( 𝑁 ∈ ℕ0 → ( 𝑁 − 0 ) = 𝑁 )
7 6 oveq2d ⊢ ( 𝑁 ∈ ℕ0 → ( 𝑁 C ( 𝑁 − 0 ) ) = ( 𝑁 C 𝑁 ) )
8 3 4 7 3eqtr3rd ⊢ ( 𝑁 ∈ ℕ0 → ( 𝑁 C 𝑁 ) = 1 )