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Metamath Proof Explorer

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 ) )