Metamath Proof Explorer


Theorem nnaddm1cl

Description: Closure of addition of positive integers minus one. (Contributed by NM, 6-Aug-2003) (Proof shortened by Mario Carneiro, 16-May-2014)

Ref Expression
Assertion nnaddm1cl ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ) → ( ( 𝐴 + 𝐵 ) − 1 ) ∈ ℕ )

Proof

Step Hyp Ref Expression
1 nncn ( 𝐴 ∈ ℕ → 𝐴 ∈ ℂ )
2 nncn ( 𝐵 ∈ ℕ → 𝐵 ∈ ℂ )
3 ax-1cn 1 ∈ ℂ
4 addsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 1 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 1 ) = ( ( 𝐴 − 1 ) + 𝐵 ) )
5 3 4 mp3an3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 1 ) = ( ( 𝐴 − 1 ) + 𝐵 ) )
6 1 2 5 syl2an ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ) → ( ( 𝐴 + 𝐵 ) − 1 ) = ( ( 𝐴 − 1 ) + 𝐵 ) )
7 nnm1nn0 ( 𝐴 ∈ ℕ → ( 𝐴 − 1 ) ∈ ℕ0 )
8 nn0nnaddcl ( ( ( 𝐴 − 1 ) ∈ ℕ0𝐵 ∈ ℕ ) → ( ( 𝐴 − 1 ) + 𝐵 ) ∈ ℕ )
9 7 8 sylan ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ) → ( ( 𝐴 − 1 ) + 𝐵 ) ∈ ℕ )
10 6 9 eqeltrd ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ) → ( ( 𝐴 + 𝐵 ) − 1 ) ∈ ℕ )