Metamath Proof Explorer


Theorem right1s

Description: The right set of 1s is empty . (Contributed by Scott Fenton, 4-Feb-2025)

Ref Expression
Assertion right1s R 1 s =

Proof

Step Hyp Ref Expression
1 rightval R 1 s = x Old bday 1 s | 1 s < s x
2 bday1s bday 1 s = 1 𝑜
3 2 fveq2i Old bday 1 s = Old 1 𝑜
4 old1 Old 1 𝑜 = 0 s
5 3 4 eqtri Old bday 1 s = 0 s
6 5 rabeqi x Old bday 1 s | 1 s < s x = x 0 s | 1 s < s x
7 breq2 x = 0 s 1 s < s x 1 s < s 0 s
8 7 rabsnif x 0 s | 1 s < s x = if 1 s < s 0 s 0 s
9 6 8 eqtri x Old bday 1 s | 1 s < s x = if 1 s < s 0 s 0 s
10 0slt1s 0 s < s 1 s
11 0sno 0 s No
12 1sno 1 s No
13 sltasym 0 s No 1 s No 0 s < s 1 s ¬ 1 s < s 0 s
14 11 12 13 mp2an 0 s < s 1 s ¬ 1 s < s 0 s
15 10 14 ax-mp ¬ 1 s < s 0 s
16 15 iffalsei if 1 s < s 0 s 0 s =
17 1 9 16 3eqtri R 1 s =