Metamath Proof Explorer


Theorem n0addscl

Description: The non-negative surreal integers are closed under addition. (Contributed by Scott Fenton, 15-Apr-2025)

Ref Expression
Assertion n0addscl ⊢ A ∈ ℕ 0s ∧ B ∈ ℕ 0s → A + s B ∈ ℕ 0s

Proof

Step Hyp Ref Expression
1 oveq2 ⊢ n = 0 s → A + s n = A + s 0 s
2 1 eleq1d ⊢ n = 0 s → A + s n ∈ ℕ 0s ↔ A + s 0 s ∈ ℕ 0s
3 2 imbi2d ⊢ n = 0 s → A ∈ ℕ 0s → A + s n ∈ ℕ 0s ↔ A ∈ ℕ 0s → A + s 0 s ∈ ℕ 0s
4 oveq2 ⊢ n = m → A + s n = A + s m
5 4 eleq1d ⊢ n = m → A + s n ∈ ℕ 0s ↔ A + s m ∈ ℕ 0s
6 5 imbi2d ⊢ n = m → A ∈ ℕ 0s → A + s n ∈ ℕ 0s ↔ A ∈ ℕ 0s → A + s m ∈ ℕ 0s
7 oveq2 ⊢ n = m + s 1 s → A + s n = A + s m + s 1 s
8 7 eleq1d ⊢ n = m + s 1 s → A + s n ∈ ℕ 0s ↔ A + s m + s 1 s ∈ ℕ 0s
9 8 imbi2d ⊢ n = m + s 1 s → A ∈ ℕ 0s → A + s n ∈ ℕ 0s ↔ A ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
10 oveq2 ⊢ n = B → A + s n = A + s B
11 10 eleq1d ⊢ n = B → A + s n ∈ ℕ 0s ↔ A + s B ∈ ℕ 0s
12 11 imbi2d ⊢ n = B → A ∈ ℕ 0s → A + s n ∈ ℕ 0s ↔ A ∈ ℕ 0s → A + s B ∈ ℕ 0s
13 n0no ⊢ A ∈ ℕ 0s → A ∈ No
14 13 addsridd ⊢ A ∈ ℕ 0s → A + s 0 s = A
15 id ⊢ A ∈ ℕ 0s → A ∈ ℕ 0s
16 14 15 eqeltrd ⊢ A ∈ ℕ 0s → A + s 0 s ∈ ℕ 0s
17 13 adantr ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s → A ∈ No
18 17 adantr ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s ∧ A + s m ∈ ℕ 0s → A ∈ No
19 n0no ⊢ m ∈ ℕ 0s → m ∈ No
20 19 adantl ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s → m ∈ No
21 20 adantr ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s ∧ A + s m ∈ ℕ 0s → m ∈ No
22 1no ⊢ 1 s ∈ No
23 22 a1i ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s ∧ A + s m ∈ ℕ 0s → 1 s ∈ No
24 18 21 23 addsassd ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s ∧ A + s m ∈ ℕ 0s → A + s m + s 1 s = A + s m + s 1 s
25 peano2n0s ⊢ A + s m ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
26 25 adantl ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s ∧ A + s m ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
27 24 26 eqeltrrd ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s ∧ A + s m ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
28 27 ex ⊢ A ∈ ℕ 0s ∧ m ∈ ℕ 0s → A + s m ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
29 28 expcom ⊢ m ∈ ℕ 0s → A ∈ ℕ 0s → A + s m ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
30 29 a2d ⊢ m ∈ ℕ 0s → A ∈ ℕ 0s → A + s m ∈ ℕ 0s → A ∈ ℕ 0s → A + s m + s 1 s ∈ ℕ 0s
31 3 6 9 12 16 30 n0sind ⊢ B ∈ ℕ 0s → A ∈ ℕ 0s → A + s B ∈ ℕ 0s
32 31 impcom ⊢ A ∈ ℕ 0s ∧ B ∈ ℕ 0s → A + s B ∈ ℕ 0s