Metamath Proof Explorer


Theorem fseqen

Description: A set that is equinumerous to its Cartesian product is equinumerous to the set of finite sequences on it. (This can be proven more easily using some choice but this proof avoids it.) (Contributed by Mario Carneiro, 18-Nov-2014)

Ref Expression
Assertion fseqen A × A A A n ω A n ω × A

Proof

Step Hyp Ref Expression
1 bren A × A A f f : A × A 1-1 onto A
2 n0 A b b A
3 exdistrv f b f : A × A 1-1 onto A b A f f : A × A 1-1 onto A b b A
4 omex ω V
5 simpl f : A × A 1-1 onto A b A f : A × A 1-1 onto A
6 f1ofo f : A × A 1-1 onto A f : A × A onto A
7 forn f : A × A onto A ran f = A
8 5 6 7 3syl f : A × A 1-1 onto A b A ran f = A
9 vex f V
10 9 rnex ran f V
11 8 10 eqeltrrdi f : A × A 1-1 onto A b A A V
12 xpexg ω V A V ω × A V
13 4 11 12 sylancr f : A × A 1-1 onto A b A ω × A V
14 simpr f : A × A 1-1 onto A b A b A
15 eqid seq ω k V , g V y A suc k g y k f y k b = seq ω k V , g V y A suc k g y k f y k b
16 eqid x n ω A n dom x seq ω k V , g V y A suc k g y k f y k b dom x x = x n ω A n dom x seq ω k V , g V y A suc k g y k f y k b dom x x
17 11 14 5 15 16 fseqenlem2 f : A × A 1-1 onto A b A x n ω A n dom x seq ω k V , g V y A suc k g y k f y k b dom x x : n ω A n 1-1 ω × A
18 f1domg ω × A V x n ω A n dom x seq ω k V , g V y A suc k g y k f y k b dom x x : n ω A n 1-1 ω × A n ω A n ω × A
19 13 17 18 sylc f : A × A 1-1 onto A b A n ω A n ω × A
20 fseqdom A V ω × A n ω A n
21 11 20 syl f : A × A 1-1 onto A b A ω × A n ω A n
22 sbth n ω A n ω × A ω × A n ω A n n ω A n ω × A
23 19 21 22 syl2anc f : A × A 1-1 onto A b A n ω A n ω × A
24 23 exlimivv f b f : A × A 1-1 onto A b A n ω A n ω × A
25 3 24 sylbir f f : A × A 1-1 onto A b b A n ω A n ω × A
26 1 2 25 syl2anb A × A A A n ω A n ω × A