Metamath Proof Explorer


Theorem alephnbtwn2

Description: No set has equinumerosity between an aleph and its successor aleph. (Contributed by NM, 3-Nov-2003) (Revised by Mario Carneiro, 2-Feb-2013)

Ref Expression
Assertion alephnbtwn2 ¬ A B B suc A

Proof

Step Hyp Ref Expression
1 cardidm card card B = card B
2 alephnbtwn card card B = card B ¬ A card B card B suc A
3 1 2 ax-mp ¬ A card B card B suc A
4 alephon suc A On
5 sdomdom B suc A B suc A
6 ondomen suc A On B suc A B dom card
7 4 5 6 sylancr B suc A B dom card
8 cardid2 B dom card card B B
9 7 8 syl B suc A card B B
10 9 ensymd B suc A B card B
11 sdomentr A B B card B A card B
12 10 11 sylan2 A B B suc A A card B
13 alephon A On
14 cardon card B On
15 onenon card B On card B dom card
16 14 15 ax-mp card B dom card
17 cardsdomel A On card B dom card A card B A card card B
18 13 16 17 mp2an A card B A card card B
19 1 eleq2i A card card B A card B
20 18 19 bitri A card B A card B
21 12 20 sylib A B B suc A A card B
22 ensdomtr card B B B suc A card B suc A
23 9 22 mpancom B suc A card B suc A
24 23 adantl A B B suc A card B suc A
25 onenon suc A On suc A dom card
26 4 25 ax-mp suc A dom card
27 cardsdomel card B On suc A dom card card B suc A card B card suc A
28 14 26 27 mp2an card B suc A card B card suc A
29 alephcard card suc A = suc A
30 29 eleq2i card B card suc A card B suc A
31 28 30 bitri card B suc A card B suc A
32 24 31 sylib A B B suc A card B suc A
33 21 32 jca A B B suc A A card B card B suc A
34 3 33 mto ¬ A B B suc A