Metamath Proof Explorer


Theorem dju1dif

Description: Adding and subtracting one gives back the original cardinality. Similar to pncan for cardinalities. (Contributed by Mario Carneiro, 18-May-2015) (Revised by Jim Kingdon, 20-Aug-2023)

Ref Expression
Assertion dju1dif AVBA⊔︀1𝑜A⊔︀1𝑜BA

Proof

Step Hyp Ref Expression
1 simpl AVBA⊔︀1𝑜AV
2 1oex 1𝑜V
3 djuex AV1𝑜VA⊔︀1𝑜V
4 1 2 3 sylancl AVBA⊔︀1𝑜A⊔︀1𝑜V
5 simpr AVBA⊔︀1𝑜BA⊔︀1𝑜
6 df1o2 1𝑜=
7 6 xpeq2i 1𝑜×1𝑜=1𝑜×
8 0ex V
9 2 8 xpsn 1𝑜×=1𝑜
10 7 9 eqtri 1𝑜×1𝑜=1𝑜
11 ssun2 1𝑜×1𝑜×A1𝑜×1𝑜
12 10 11 eqsstrri 1𝑜×A1𝑜×1𝑜
13 opex 1𝑜V
14 13 snss 1𝑜×A1𝑜×1𝑜1𝑜×A1𝑜×1𝑜
15 12 14 mpbir 1𝑜×A1𝑜×1𝑜
16 df-dju A⊔︀1𝑜=×A1𝑜×1𝑜
17 15 16 eleqtrri 1𝑜A⊔︀1𝑜
18 17 a1i AVBA⊔︀1𝑜1𝑜A⊔︀1𝑜
19 difsnen A⊔︀1𝑜VBA⊔︀1𝑜1𝑜A⊔︀1𝑜A⊔︀1𝑜BA⊔︀1𝑜1𝑜
20 4 5 18 19 syl3anc AVBA⊔︀1𝑜A⊔︀1𝑜BA⊔︀1𝑜1𝑜
21 16 difeq1i A⊔︀1𝑜1𝑜=×A1𝑜×1𝑜1𝑜
22 xp01disjl ×A1𝑜×1𝑜=
23 disj3 ×A1𝑜×1𝑜=×A=×A1𝑜×1𝑜
24 22 23 mpbi ×A=×A1𝑜×1𝑜
25 difun2 ×A1𝑜×1𝑜1𝑜×1𝑜=×A1𝑜×1𝑜
26 10 difeq2i ×A1𝑜×1𝑜1𝑜×1𝑜=×A1𝑜×1𝑜1𝑜
27 24 25 26 3eqtr2i ×A=×A1𝑜×1𝑜1𝑜
28 21 27 eqtr4i A⊔︀1𝑜1𝑜=×A
29 xpsnen2g VAV×AA
30 8 1 29 sylancr AVBA⊔︀1𝑜×AA
31 28 30 eqbrtrid AVBA⊔︀1𝑜A⊔︀1𝑜1𝑜A
32 entr A⊔︀1𝑜BA⊔︀1𝑜1𝑜A⊔︀1𝑜1𝑜AA⊔︀1𝑜BA
33 20 31 32 syl2anc AVBA⊔︀1𝑜A⊔︀1𝑜BA