Metamath Proof Explorer


Theorem n0s0suc

Description: A non-negative surreal integer is either zero or a successor. (Contributed by Scott Fenton, 26-Jul-2025)

Ref Expression
Assertion n0s0suc ⊢ A ∈ ℕ 0s → A = 0 s ∨ ∃ x ∈ ℕ 0s A = x + s 1 s

Proof

Step Hyp Ref Expression
1 eqeq1 ⊢ y = 0 s → y = 0 s ↔ 0 s = 0 s
2 eqeq1 ⊢ y = 0 s → y = x + s 1 s ↔ 0 s = x + s 1 s
3 2 rexbidv ⊢ y = 0 s → ∃ x ∈ ℕ 0s y = x + s 1 s ↔ ∃ x ∈ ℕ 0s 0 s = x + s 1 s
4 1 3 orbi12d ⊢ y = 0 s → y = 0 s ∨ ∃ x ∈ ℕ 0s y = x + s 1 s ↔ 0 s = 0 s ∨ ∃ x ∈ ℕ 0s 0 s = x + s 1 s
5 eqeq1 ⊢ y = z → y = 0 s ↔ z = 0 s
6 eqeq1 ⊢ y = z → y = x + s 1 s ↔ z = x + s 1 s
7 6 rexbidv ⊢ y = z → ∃ x ∈ ℕ 0s y = x + s 1 s ↔ ∃ x ∈ ℕ 0s z = x + s 1 s
8 5 7 orbi12d ⊢ y = z → y = 0 s ∨ ∃ x ∈ ℕ 0s y = x + s 1 s ↔ z = 0 s ∨ ∃ x ∈ ℕ 0s z = x + s 1 s
9 eqeq1 ⊢ y = z + s 1 s → y = 0 s ↔ z + s 1 s = 0 s
10 eqeq1 ⊢ y = z + s 1 s → y = x + s 1 s ↔ z + s 1 s = x + s 1 s
11 10 rexbidv ⊢ y = z + s 1 s → ∃ x ∈ ℕ 0s y = x + s 1 s ↔ ∃ x ∈ ℕ 0s z + s 1 s = x + s 1 s
12 9 11 orbi12d ⊢ y = z + s 1 s → y = 0 s ∨ ∃ x ∈ ℕ 0s y = x + s 1 s ↔ z + s 1 s = 0 s ∨ ∃ x ∈ ℕ 0s z + s 1 s = x + s 1 s
13 eqeq1 ⊢ y = A → y = 0 s ↔ A = 0 s
14 eqeq1 ⊢ y = A → y = x + s 1 s ↔ A = x + s 1 s
15 14 rexbidv ⊢ y = A → ∃ x ∈ ℕ 0s y = x + s 1 s ↔ ∃ x ∈ ℕ 0s A = x + s 1 s
16 13 15 orbi12d ⊢ y = A → y = 0 s ∨ ∃ x ∈ ℕ 0s y = x + s 1 s ↔ A = 0 s ∨ ∃ x ∈ ℕ 0s A = x + s 1 s
17 eqid ⊢ 0 s = 0 s
18 17 orci ⊢ 0 s = 0 s ∨ ∃ x ∈ ℕ 0s 0 s = x + s 1 s
19 clel5 ⊢ z ∈ ℕ 0s ↔ ∃ x ∈ ℕ 0s z = x
20 19 biimpi ⊢ z ∈ ℕ 0s → ∃ x ∈ ℕ 0s z = x
21 n0no ⊢ z ∈ ℕ 0s → z ∈ No
22 n0no ⊢ x ∈ ℕ 0s → x ∈ No
23 1no ⊢ 1 s ∈ No
24 addscan2 ⊢ z ∈ No ∧ x ∈ No ∧ 1 s ∈ No → z + s 1 s = x + s 1 s ↔ z = x
25 23 24 mp3an3 ⊢ z ∈ No ∧ x ∈ No → z + s 1 s = x + s 1 s ↔ z = x
26 21 22 25 syl2an ⊢ z ∈ ℕ 0s ∧ x ∈ ℕ 0s → z + s 1 s = x + s 1 s ↔ z = x
27 26 rexbidva ⊢ z ∈ ℕ 0s → ∃ x ∈ ℕ 0s z + s 1 s = x + s 1 s ↔ ∃ x ∈ ℕ 0s z = x
28 20 27 mpbird ⊢ z ∈ ℕ 0s → ∃ x ∈ ℕ 0s z + s 1 s = x + s 1 s
29 28 olcd ⊢ z ∈ ℕ 0s → z + s 1 s = 0 s ∨ ∃ x ∈ ℕ 0s z + s 1 s = x + s 1 s
30 29 a1d ⊢ z ∈ ℕ 0s → z = 0 s ∨ ∃ x ∈ ℕ 0s z = x + s 1 s → z + s 1 s = 0 s ∨ ∃ x ∈ ℕ 0s z + s 1 s = x + s 1 s
31 4 8 12 16 18 30 n0sind ⊢ A ∈ ℕ 0s → A = 0 s ∨ ∃ x ∈ ℕ 0s A = x + s 1 s