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 0s A = 0 s x 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 0s y = x + s 1 s x 0s 0 s = x + s 1 s
4 1 3 orbi12d y = 0 s y = 0 s x 0s y = x + s 1 s 0 s = 0 s x 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 0s y = x + s 1 s x 0s z = x + s 1 s
8 5 7 orbi12d y = z y = 0 s x 0s y = x + s 1 s z = 0 s x 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 0s y = x + s 1 s x 0s z + s 1 s = x + s 1 s
12 9 11 orbi12d y = z + s 1 s y = 0 s x 0s y = x + s 1 s z + s 1 s = 0 s x 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 0s y = x + s 1 s x 0s A = x + s 1 s
16 13 15 orbi12d y = A y = 0 s x 0s y = x + s 1 s A = 0 s x 0s A = x + s 1 s
17 eqid 0 s = 0 s
18 17 orci 0 s = 0 s x 0s 0 s = x + s 1 s
19 clel5 z 0s x 0s z = x
20 19 biimpi z 0s x 0s z = x
21 n0sno z 0s z No
22 n0sno x 0s x No
23 1sno 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 0s x 0s z + s 1 s = x + s 1 s z = x
27 26 rexbidva z 0s x 0s z + s 1 s = x + s 1 s x 0s z = x
28 20 27 mpbird z 0s x 0s z + s 1 s = x + s 1 s
29 28 olcd z 0s z + s 1 s = 0 s x 0s z + s 1 s = x + s 1 s
30 29 a1d z 0s z = 0 s x 0s z = x + s 1 s z + s 1 s = 0 s x 0s z + s 1 s = x + s 1 s
31 4 8 12 16 18 30 n0sind A 0s A = 0 s x 0s A = x + s 1 s