Metamath Proof Explorer


Theorem naddoa

Description: Natural addition of a natural is the same as regular addition. (Contributed by Scott Fenton, 20-Aug-2025)

Ref Expression
Assertion naddoa ⊢ A ∈ On ∧ B ∈ ω → A + B = A + 𝑜 B

Proof

Step Hyp Ref Expression
1 oveq2 ⊢ y = ∅ → A + y = A + ∅
2 oveq2 ⊢ y = ∅ → A + 𝑜 y = A + 𝑜 ∅
3 1 2 eqeq12d ⊢ y = ∅ → A + y = A + 𝑜 y ↔ A + ∅ = A + 𝑜 ∅
4 3 imbi2d ⊢ y = ∅ → A ∈ On → A + y = A + 𝑜 y ↔ A ∈ On → A + ∅ = A + 𝑜 ∅
5 oveq2 ⊢ y = x → A + y = A + x
6 oveq2 ⊢ y = x → A + 𝑜 y = A + 𝑜 x
7 5 6 eqeq12d ⊢ y = x → A + y = A + 𝑜 y ↔ A + x = A + 𝑜 x
8 7 imbi2d ⊢ y = x → A ∈ On → A + y = A + 𝑜 y ↔ A ∈ On → A + x = A + 𝑜 x
9 oveq2 ⊢ y = suc ⁡ x → A + y = A + suc ⁡ x
10 oveq2 ⊢ y = suc ⁡ x → A + 𝑜 y = A + 𝑜 suc ⁡ x
11 9 10 eqeq12d ⊢ y = suc ⁡ x → A + y = A + 𝑜 y ↔ A + suc ⁡ x = A + 𝑜 suc ⁡ x
12 11 imbi2d ⊢ y = suc ⁡ x → A ∈ On → A + y = A + 𝑜 y ↔ A ∈ On → A + suc ⁡ x = A + 𝑜 suc ⁡ x
13 oveq2 ⊢ y = B → A + y = A + B
14 oveq2 ⊢ y = B → A + 𝑜 y = A + 𝑜 B
15 13 14 eqeq12d ⊢ y = B → A + y = A + 𝑜 y ↔ A + B = A + 𝑜 B
16 15 imbi2d ⊢ y = B → A ∈ On → A + y = A + 𝑜 y ↔ A ∈ On → A + B = A + 𝑜 B
17 naddrid ⊢ A ∈ On → A + ∅ = A
18 oa0 ⊢ A ∈ On → A + 𝑜 ∅ = A
19 17 18 eqtr4d ⊢ A ∈ On → A + ∅ = A + 𝑜 ∅
20 suceq ⊢ A + x = A + 𝑜 x → suc ⁡ A + x = suc ⁡ A + 𝑜 x
21 20 3ad2ant3 ⊢ x ∈ ω ∧ A ∈ On ∧ A + x = A + 𝑜 x → suc ⁡ A + x = suc ⁡ A + 𝑜 x
22 nnon ⊢ x ∈ ω → x ∈ On
23 naddsuc2 ⊢ A ∈ On ∧ x ∈ On → A + suc ⁡ x = suc ⁡ A + x
24 22 23 sylan2 ⊢ A ∈ On ∧ x ∈ ω → A + suc ⁡ x = suc ⁡ A + x
25 24 ancoms ⊢ x ∈ ω ∧ A ∈ On → A + suc ⁡ x = suc ⁡ A + x
26 25 3adant3 ⊢ x ∈ ω ∧ A ∈ On ∧ A + x = A + 𝑜 x → A + suc ⁡ x = suc ⁡ A + x
27 onasuc ⊢ A ∈ On ∧ x ∈ ω → A + 𝑜 suc ⁡ x = suc ⁡ A + 𝑜 x
28 27 ancoms ⊢ x ∈ ω ∧ A ∈ On → A + 𝑜 suc ⁡ x = suc ⁡ A + 𝑜 x
29 28 3adant3 ⊢ x ∈ ω ∧ A ∈ On ∧ A + x = A + 𝑜 x → A + 𝑜 suc ⁡ x = suc ⁡ A + 𝑜 x
30 21 26 29 3eqtr4d ⊢ x ∈ ω ∧ A ∈ On ∧ A + x = A + 𝑜 x → A + suc ⁡ x = A + 𝑜 suc ⁡ x
31 30 3exp ⊢ x ∈ ω → A ∈ On → A + x = A + 𝑜 x → A + suc ⁡ x = A + 𝑜 suc ⁡ x
32 31 a2d ⊢ x ∈ ω → A ∈ On → A + x = A + 𝑜 x → A ∈ On → A + suc ⁡ x = A + 𝑜 suc ⁡ x
33 4 8 12 16 19 32 finds ⊢ B ∈ ω → A ∈ On → A + B = A + 𝑜 B
34 33 impcom ⊢ A ∈ On ∧ B ∈ ω → A + B = A + 𝑜 B