Theorem nnadd1com

Description: Addition with 1 is commutative for natural numbers. (Contributed by Steven Nguyen, 9-Dec-2022)

	Ref	Expression
Assertion	nnadd1com	⊢ ( 𝐴 ∈ ℕ → ( 𝐴 + 1 ) = ( 1 + 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		oveq1	⊢ ( 𝑥 = 1 → ( 𝑥 + 1 ) = ( 1 + 1 ) )
2		oveq2	⊢ ( 𝑥 = 1 → ( 1 + 𝑥 ) = ( 1 + 1 ) )
3	1 2	eqeq12d	⊢ ( 𝑥 = 1 → ( ( 𝑥 + 1 ) = ( 1 + 𝑥 ) ↔ ( 1 + 1 ) = ( 1 + 1 ) ) )
4		oveq1	⊢ ( 𝑥 = 𝑦 → ( 𝑥 + 1 ) = ( 𝑦 + 1 ) )
5		oveq2	⊢ ( 𝑥 = 𝑦 → ( 1 + 𝑥 ) = ( 1 + 𝑦 ) )
6	4 5	eqeq12d	⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 + 1 ) = ( 1 + 𝑥 ) ↔ ( 𝑦 + 1 ) = ( 1 + 𝑦 ) ) )
7		oveq1	⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( 𝑥 + 1 ) = ( ( 𝑦 + 1 ) + 1 ) )
8		oveq2	⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( 1 + 𝑥 ) = ( 1 + ( 𝑦 + 1 ) ) )
9	7 8	eqeq12d	⊢ ( 𝑥 = ( 𝑦 + 1 ) → ( ( 𝑥 + 1 ) = ( 1 + 𝑥 ) ↔ ( ( 𝑦 + 1 ) + 1 ) = ( 1 + ( 𝑦 + 1 ) ) ) )
10		oveq1	⊢ ( 𝑥 = 𝐴 → ( 𝑥 + 1 ) = ( 𝐴 + 1 ) )
11		oveq2	⊢ ( 𝑥 = 𝐴 → ( 1 + 𝑥 ) = ( 1 + 𝐴 ) )
12	10 11	eqeq12d	⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 + 1 ) = ( 1 + 𝑥 ) ↔ ( 𝐴 + 1 ) = ( 1 + 𝐴 ) ) )
13		eqid	⊢ ( 1 + 1 ) = ( 1 + 1 )
14		oveq1	⊢ ( ( 𝑦 + 1 ) = ( 1 + 𝑦 ) → ( ( 𝑦 + 1 ) + 1 ) = ( ( 1 + 𝑦 ) + 1 ) )
15		1cnd	⊢ ( 𝑦 ∈ ℕ → 1 ∈ ℂ )
16		nncn	⊢ ( 𝑦 ∈ ℕ → 𝑦 ∈ ℂ )
17	15 16 15	addassd	⊢ ( 𝑦 ∈ ℕ → ( ( 1 + 𝑦 ) + 1 ) = ( 1 + ( 𝑦 + 1 ) ) )
18	14 17	sylan9eqr	⊢ ( ( 𝑦 ∈ ℕ ∧ ( 𝑦 + 1 ) = ( 1 + 𝑦 ) ) → ( ( 𝑦 + 1 ) + 1 ) = ( 1 + ( 𝑦 + 1 ) ) )
19	18	ex	⊢ ( 𝑦 ∈ ℕ → ( ( 𝑦 + 1 ) = ( 1 + 𝑦 ) → ( ( 𝑦 + 1 ) + 1 ) = ( 1 + ( 𝑦 + 1 ) ) ) )
20	3 6 9 12 13 19	nnind	⊢ ( 𝐴 ∈ ℕ → ( 𝐴 + 1 ) = ( 1 + 𝐴 ) )