Metamath Proof Explorer


Theorem nnadju

Description: The cardinal and ordinal sums of finite ordinals are equal. For a shorter proof using ax-rep , see nnadjuALT . (Contributed by Paul Chapman, 11-Apr-2009) (Revised by Mario Carneiro, 6-Feb-2013) Avoid ax-rep . (Revised by BTernaryTau, 2-Jul-2024)

Ref Expression
Assertion nnadju AωBωcardA⊔︀B=A+𝑜B

Proof

Step Hyp Ref Expression
1 djueq2 x=BA⊔︀x=A⊔︀B
2 oveq2 x=BA+𝑜x=A+𝑜B
3 1 2 breq12d x=BA⊔︀xA+𝑜xA⊔︀BA+𝑜B
4 3 imbi2d x=BAωA⊔︀xA+𝑜xAωA⊔︀BA+𝑜B
5 djueq2 x=A⊔︀x=A⊔︀
6 oveq2 x=A+𝑜x=A+𝑜
7 5 6 breq12d x=A⊔︀xA+𝑜xA⊔︀A+𝑜
8 djueq2 x=yA⊔︀x=A⊔︀y
9 oveq2 x=yA+𝑜x=A+𝑜y
10 8 9 breq12d x=yA⊔︀xA+𝑜xA⊔︀yA+𝑜y
11 djueq2 x=sucyA⊔︀x=A⊔︀sucy
12 oveq2 x=sucyA+𝑜x=A+𝑜sucy
13 11 12 breq12d x=sucyA⊔︀xA+𝑜xA⊔︀sucyA+𝑜sucy
14 dju0en AωA⊔︀A
15 nna0 AωA+𝑜=A
16 14 15 breqtrrd AωA⊔︀A+𝑜
17 1oex 1𝑜V
18 djuassen Aωyω1𝑜VA⊔︀y⊔︀1𝑜A⊔︀y⊔︀1𝑜
19 17 18 mp3an3 AωyωA⊔︀y⊔︀1𝑜A⊔︀y⊔︀1𝑜
20 enrefg AωAA
21 nnord yωOrdy
22 ordirr Ordy¬yy
23 21 22 syl yω¬yy
24 dju1en yω¬yyy⊔︀1𝑜sucy
25 23 24 mpdan yωy⊔︀1𝑜sucy
26 djuen AAy⊔︀1𝑜sucyA⊔︀y⊔︀1𝑜A⊔︀sucy
27 20 25 26 syl2an AωyωA⊔︀y⊔︀1𝑜A⊔︀sucy
28 entr A⊔︀y⊔︀1𝑜A⊔︀y⊔︀1𝑜A⊔︀y⊔︀1𝑜A⊔︀sucyA⊔︀y⊔︀1𝑜A⊔︀sucy
29 19 27 28 syl2anc AωyωA⊔︀y⊔︀1𝑜A⊔︀sucy
30 29 ensymd AωyωA⊔︀sucyA⊔︀y⊔︀1𝑜
31 17 enref 1𝑜1𝑜
32 djuen A⊔︀yA+𝑜y1𝑜1𝑜A⊔︀y⊔︀1𝑜A+𝑜y⊔︀1𝑜
33 31 32 mpan2 A⊔︀yA+𝑜yA⊔︀y⊔︀1𝑜A+𝑜y⊔︀1𝑜
34 33 a1i AωyωA⊔︀yA+𝑜yA⊔︀y⊔︀1𝑜A+𝑜y⊔︀1𝑜
35 nnacl AωyωA+𝑜yω
36 nnord A+𝑜yωOrdA+𝑜y
37 ordirr OrdA+𝑜y¬A+𝑜yA+𝑜y
38 35 36 37 3syl Aωyω¬A+𝑜yA+𝑜y
39 dju1en A+𝑜yω¬A+𝑜yA+𝑜yA+𝑜y⊔︀1𝑜sucA+𝑜y
40 35 38 39 syl2anc AωyωA+𝑜y⊔︀1𝑜sucA+𝑜y
41 nnasuc AωyωA+𝑜sucy=sucA+𝑜y
42 40 41 breqtrrd AωyωA+𝑜y⊔︀1𝑜A+𝑜sucy
43 34 42 jctird AωyωA⊔︀yA+𝑜yA⊔︀y⊔︀1𝑜A+𝑜y⊔︀1𝑜A+𝑜y⊔︀1𝑜A+𝑜sucy
44 entr A⊔︀y⊔︀1𝑜A+𝑜y⊔︀1𝑜A+𝑜y⊔︀1𝑜A+𝑜sucyA⊔︀y⊔︀1𝑜A+𝑜sucy
45 43 44 syl6 AωyωA⊔︀yA+𝑜yA⊔︀y⊔︀1𝑜A+𝑜sucy
46 entr A⊔︀sucyA⊔︀y⊔︀1𝑜A⊔︀y⊔︀1𝑜A+𝑜sucyA⊔︀sucyA+𝑜sucy
47 30 45 46 syl6an AωyωA⊔︀yA+𝑜yA⊔︀sucyA+𝑜sucy
48 47 expcom yωAωA⊔︀yA+𝑜yA⊔︀sucyA+𝑜sucy
49 7 10 13 16 48 finds2 xωAωA⊔︀xA+𝑜x
50 4 49 vtoclga BωAωA⊔︀BA+𝑜B
51 50 impcom AωBωA⊔︀BA+𝑜B
52 carden2b A⊔︀BA+𝑜BcardA⊔︀B=cardA+𝑜B
53 51 52 syl AωBωcardA⊔︀B=cardA+𝑜B
54 nnacl AωBωA+𝑜Bω
55 cardnn A+𝑜BωcardA+𝑜B=A+𝑜B
56 54 55 syl AωBωcardA+𝑜B=A+𝑜B
57 53 56 eqtrd AωBωcardA⊔︀B=A+𝑜B