Metamath Proof Explorer
		
		
		
		Description:  Adding a multiple of one operand of the gcd operator to the other does
       not alter the result.  (Contributed by metakunt, 25-Apr-2024)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypotheses | gcdaddmzz2nni.1 | ⊢ 𝑀  ∈  ℕ | 
					
						|  |  | gcdaddmzz2nni.2 | ⊢ 𝑁  ∈  ℕ | 
					
						|  |  | gcdaddmzz2nni.3 | ⊢ 𝐾  ∈  ℤ | 
				
					|  | Assertion | gcdaddmzz2nni | ⊢  ( 𝑀  gcd  𝑁 )  =  ( 𝑀  gcd  ( 𝑁  +  ( 𝐾  ·  𝑀 ) ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | gcdaddmzz2nni.1 | ⊢ 𝑀  ∈  ℕ | 
						
							| 2 |  | gcdaddmzz2nni.2 | ⊢ 𝑁  ∈  ℕ | 
						
							| 3 |  | gcdaddmzz2nni.3 | ⊢ 𝐾  ∈  ℤ | 
						
							| 4 | 1 | nnzi | ⊢ 𝑀  ∈  ℤ | 
						
							| 5 | 2 | nnzi | ⊢ 𝑁  ∈  ℤ | 
						
							| 6 | 3 4 5 | 3pm3.2i | ⊢ ( 𝐾  ∈  ℤ  ∧  𝑀  ∈  ℤ  ∧  𝑁  ∈  ℤ ) | 
						
							| 7 |  | gcdaddm | ⊢ ( ( 𝐾  ∈  ℤ  ∧  𝑀  ∈  ℤ  ∧  𝑁  ∈  ℤ )  →  ( 𝑀  gcd  𝑁 )  =  ( 𝑀  gcd  ( 𝑁  +  ( 𝐾  ·  𝑀 ) ) ) ) | 
						
							| 8 | 6 7 | ax-mp | ⊢ ( 𝑀  gcd  𝑁 )  =  ( 𝑀  gcd  ( 𝑁  +  ( 𝐾  ·  𝑀 ) ) ) |