Metamath Proof Explorer


Theorem congid

Description: Every integer is congruent to itself mod every base. (Contributed by Stefan O'Rear, 1-Oct-2014)

Ref Expression
Assertion congid ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → 𝐴 ∥ ( 𝐵𝐵 ) )

Proof

Step Hyp Ref Expression
1 dvds0 ( 𝐴 ∈ ℤ → 𝐴 ∥ 0 )
2 1 adantr ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → 𝐴 ∥ 0 )
3 zcn ( 𝐵 ∈ ℤ → 𝐵 ∈ ℂ )
4 3 adantl ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → 𝐵 ∈ ℂ )
5 4 subidd ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( 𝐵𝐵 ) = 0 )
6 2 5 breqtrrd ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → 𝐴 ∥ ( 𝐵𝐵 ) )