Theorem dvdsgcdidd

Description: The greatest common divisor of a positive integer and another integer it divides is itself. (Contributed by Rohan Ridenour, 3-Aug-2023)

	Ref	Expression
Hypotheses	dvdsgcdidd.1	⊢ ( 𝜑 → 𝑀 ∈ ℕ )
	dvdsgcdidd.2	⊢ ( 𝜑 → 𝑁 ∈ ℤ )
	dvdsgcdidd.3	⊢ ( 𝜑 → 𝑀 ∥ 𝑁 )
Assertion	dvdsgcdidd	⊢ ( 𝜑 → ( 𝑀 gcd 𝑁 ) = 𝑀 )

Proof

Step	Hyp	Ref	Expression
1		dvdsgcdidd.1	⊢ ( 𝜑 → 𝑀 ∈ ℕ )
2		dvdsgcdidd.2	⊢ ( 𝜑 → 𝑁 ∈ ℤ )
3		dvdsgcdidd.3	⊢ ( 𝜑 → 𝑀 ∥ 𝑁 )
4	2	zcnd	⊢ ( 𝜑 → 𝑁 ∈ ℂ )
5	1	nncnd	⊢ ( 𝜑 → 𝑀 ∈ ℂ )
6	1	nnne0d	⊢ ( 𝜑 → 𝑀 ≠ 0 )
7	4 5 6	divcan1d	⊢ ( 𝜑 → ( ( 𝑁 / 𝑀 ) · 𝑀 ) = 𝑁 )
8	7	oveq2d	⊢ ( 𝜑 → ( 𝑀 gcd ( ( 𝑁 / 𝑀 ) · 𝑀 ) ) = ( 𝑀 gcd 𝑁 ) )
9	1	nnnn0d	⊢ ( 𝜑 → 𝑀 ∈ ℕ0 )
10	1	nnzd	⊢ ( 𝜑 → 𝑀 ∈ ℤ )
11		dvdsval2	⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑀 ≠ 0 ∧ 𝑁 ∈ ℤ ) → ( 𝑀 ∥ 𝑁 ↔ ( 𝑁 / 𝑀 ) ∈ ℤ ) )
12	10 6 2 11	syl3anc	⊢ ( 𝜑 → ( 𝑀 ∥ 𝑁 ↔ ( 𝑁 / 𝑀 ) ∈ ℤ ) )
13	3 12	mpbid	⊢ ( 𝜑 → ( 𝑁 / 𝑀 ) ∈ ℤ )
14	9 13	gcdmultipled	⊢ ( 𝜑 → ( 𝑀 gcd ( ( 𝑁 / 𝑀 ) · 𝑀 ) ) = 𝑀 )
15	8 14	eqtr3d	⊢ ( 𝜑 → ( 𝑀 gcd 𝑁 ) = 𝑀 )