Metamath Proof Explorer


Theorem naddwordnexlem0

Description: When A is the sum of a limit ordinal (or zero) and a natural number and B is the sum of a larger limit ordinal and a smaller natural number, ( _om .o suc C ) lies between A and B . (Contributed by RP, 14-Feb-2025)

Ref Expression
Hypotheses naddwordnex.a φ A = ω 𝑜 C + 𝑜 M
naddwordnex.b φ B = ω 𝑜 D + 𝑜 N
naddwordnex.c φ C D
naddwordnex.d φ D On
naddwordnex.m φ M ω
naddwordnex.n φ N M
Assertion naddwordnexlem0 φ A ω 𝑜 suc C ω 𝑜 suc C B

Proof

Step Hyp Ref Expression
1 naddwordnex.a φ A = ω 𝑜 C + 𝑜 M
2 naddwordnex.b φ B = ω 𝑜 D + 𝑜 N
3 naddwordnex.c φ C D
4 naddwordnex.d φ D On
5 naddwordnex.m φ M ω
6 naddwordnex.n φ N M
7 omelon ω On
8 7 a1i φ ω On
9 onelon D On C D C On
10 4 3 9 syl2anc φ C On
11 omcl ω On C On ω 𝑜 C On
12 8 10 11 syl2anc φ ω 𝑜 C On
13 8 12 jca φ ω On ω 𝑜 C On
14 oaordi ω On ω 𝑜 C On M ω ω 𝑜 C + 𝑜 M ω 𝑜 C + 𝑜 ω
15 13 5 14 sylc φ ω 𝑜 C + 𝑜 M ω 𝑜 C + 𝑜 ω
16 omsuc ω On C On ω 𝑜 suc C = ω 𝑜 C + 𝑜 ω
17 8 10 16 syl2anc φ ω 𝑜 suc C = ω 𝑜 C + 𝑜 ω
18 15 1 17 3eltr4d φ A ω 𝑜 suc C
19 onsuc C On suc C On
20 10 19 syl φ suc C On
21 20 4 8 3jca φ suc C On D On ω On
22 onsucss D On C D suc C D
23 4 3 22 sylc φ suc C D
24 omwordi suc C On D On ω On suc C D ω 𝑜 suc C ω 𝑜 D
25 21 23 24 sylc φ ω 𝑜 suc C ω 𝑜 D
26 omcl ω On D On ω 𝑜 D On
27 8 4 26 syl2anc φ ω 𝑜 D On
28 6 5 jca φ N M M ω
29 ontr1 ω On N M M ω N ω
30 8 28 29 sylc φ N ω
31 nnon N ω N On
32 30 31 syl φ N On
33 oaword1 ω 𝑜 D On N On ω 𝑜 D ω 𝑜 D + 𝑜 N
34 27 32 33 syl2anc φ ω 𝑜 D ω 𝑜 D + 𝑜 N
35 25 34 sstrd φ ω 𝑜 suc C ω 𝑜 D + 𝑜 N
36 35 2 sseqtrrd φ ω 𝑜 suc C B
37 18 36 jca φ A ω 𝑜 suc C ω 𝑜 suc C B