Metamath Proof Explorer


Theorem qnnen

Description: The rational numbers are countable. This proof does not use the Axiom of Choice, even though it uses an onto function, because the base set ( ZZ X. NN ) is numerable. Exercise 2 of Enderton p. 133. For purposes of the Metamath 100 list, we are considering Mario Carneiro's revision as the date this proof was completed. This is Metamath 100 proof #3. (Contributed by NM, 31-Jul-2004) (Revised by Mario Carneiro, 3-Mar-2013)

Ref Expression
Assertion qnnen

Proof

Step Hyp Ref Expression
1 omelon ωOn
2 nnenom ω
3 2 ensymi ω
4 isnumi ωOnωdomcard
5 1 3 4 mp2an domcard
6 znnen
7 ennum domcarddomcard
8 6 7 ax-mp domcarddomcard
9 5 8 mpbir domcard
10 xpnum domcarddomcard×domcard
11 9 5 10 mp2an ×domcard
12 eqid x,yxy=x,yxy
13 ovex xyV
14 12 13 fnmpoi x,yxyFn×
15 12 rnmpo ranx,yxy=z|xyz=xy
16 elq zxyz=xy
17 16 eqabi =z|xyz=xy
18 15 17 eqtr4i ranx,yxy=
19 df-fo x,yxy:×ontox,yxyFn×ranx,yxy=
20 14 18 19 mpbir2an x,yxy:×onto
21 fodomnum ×domcardx,yxy:×onto×
22 11 20 21 mp2 ×
23 nnex V
24 23 enref
25 xpen ××
26 6 24 25 mp2an ××
27 xpnnen ×
28 26 27 entri ×
29 domentr ××
30 22 28 29 mp2an
31 qex V
32 nnssq
33 ssdomg V
34 31 32 33 mp2
35 sbth
36 30 34 35 mp2an