Metamath Proof Explorer


Theorem oacomf1o

Description: Define a bijection from A +o B to B +o A . Thus, the two are equinumerous even if they are not equal (which sometimes occurs, e.g., oancom ). (Contributed by Mario Carneiro, 30-May-2015)

Ref Expression
Hypothesis oacomf1o.1 F=xAB+𝑜xxBA+𝑜x-1
Assertion oacomf1o AOnBOnF:A+𝑜B1-1 ontoB+𝑜A

Proof

Step Hyp Ref Expression
1 oacomf1o.1 F=xAB+𝑜xxBA+𝑜x-1
2 eqid xAB+𝑜x=xAB+𝑜x
3 2 oacomf1olem AOnBOnxAB+𝑜x:A1-1 ontoranxAB+𝑜xranxAB+𝑜xB=
4 3 simpld AOnBOnxAB+𝑜x:A1-1 ontoranxAB+𝑜x
5 eqid xBA+𝑜x=xBA+𝑜x
6 5 oacomf1olem BOnAOnxBA+𝑜x:B1-1 ontoranxBA+𝑜xranxBA+𝑜xA=
7 6 ancoms AOnBOnxBA+𝑜x:B1-1 ontoranxBA+𝑜xranxBA+𝑜xA=
8 7 simpld AOnBOnxBA+𝑜x:B1-1 ontoranxBA+𝑜x
9 f1ocnv xBA+𝑜x:B1-1 ontoranxBA+𝑜xxBA+𝑜x-1:ranxBA+𝑜x1-1 ontoB
10 8 9 syl AOnBOnxBA+𝑜x-1:ranxBA+𝑜x1-1 ontoB
11 incom AranxBA+𝑜x=ranxBA+𝑜xA
12 7 simprd AOnBOnranxBA+𝑜xA=
13 11 12 eqtrid AOnBOnAranxBA+𝑜x=
14 3 simprd AOnBOnranxAB+𝑜xB=
15 f1oun xAB+𝑜x:A1-1 ontoranxAB+𝑜xxBA+𝑜x-1:ranxBA+𝑜x1-1 ontoBAranxBA+𝑜x=ranxAB+𝑜xB=xAB+𝑜xxBA+𝑜x-1:AranxBA+𝑜x1-1 ontoranxAB+𝑜xB
16 4 10 13 14 15 syl22anc AOnBOnxAB+𝑜xxBA+𝑜x-1:AranxBA+𝑜x1-1 ontoranxAB+𝑜xB
17 f1oeq1 F=xAB+𝑜xxBA+𝑜x-1F:AranxBA+𝑜x1-1 ontoranxAB+𝑜xBxAB+𝑜xxBA+𝑜x-1:AranxBA+𝑜x1-1 ontoranxAB+𝑜xB
18 1 17 ax-mp F:AranxBA+𝑜x1-1 ontoranxAB+𝑜xBxAB+𝑜xxBA+𝑜x-1:AranxBA+𝑜x1-1 ontoranxAB+𝑜xB
19 16 18 sylibr AOnBOnF:AranxBA+𝑜x1-1 ontoranxAB+𝑜xB
20 oarec AOnBOnA+𝑜B=AranxBA+𝑜x
21 20 f1oeq2d AOnBOnF:A+𝑜B1-1 ontoranxAB+𝑜xBF:AranxBA+𝑜x1-1 ontoranxAB+𝑜xB
22 19 21 mpbird AOnBOnF:A+𝑜B1-1 ontoranxAB+𝑜xB
23 oarec BOnAOnB+𝑜A=BranxAB+𝑜x
24 23 ancoms AOnBOnB+𝑜A=BranxAB+𝑜x
25 uncom BranxAB+𝑜x=ranxAB+𝑜xB
26 24 25 eqtrdi AOnBOnB+𝑜A=ranxAB+𝑜xB
27 26 f1oeq3d AOnBOnF:A+𝑜B1-1 ontoB+𝑜AF:A+𝑜B1-1 ontoranxAB+𝑜xB
28 22 27 mpbird AOnBOnF:A+𝑜B1-1 ontoB+𝑜A