Metamath Proof Explorer


Theorem isismty

Description: The condition "is an isometry". (Contributed by Jeff Madsen, 2-Sep-2009)

Ref Expression
Assertion isismty M∞MetXN∞MetYFMIsmtyNF:X1-1 ontoYxXyXxMy=FxNFy

Proof

Step Hyp Ref Expression
1 ismtyval M∞MetXN∞MetYMIsmtyN=f|f:X1-1 ontoYxXyXxMy=fxNfy
2 1 eleq2d M∞MetXN∞MetYFMIsmtyNFf|f:X1-1 ontoYxXyXxMy=fxNfy
3 f1of F:X1-1 ontoYF:XY
4 3 adantr F:X1-1 ontoYxXyXxMy=FxNFyF:XY
5 elfvdm M∞MetXXdom∞Met
6 elfvdm N∞MetYYdom∞Met
7 fex2 F:XYXdom∞MetYdom∞MetFV
8 4 5 6 7 syl3an F:X1-1 ontoYxXyXxMy=FxNFyM∞MetXN∞MetYFV
9 8 3expib F:X1-1 ontoYxXyXxMy=FxNFyM∞MetXN∞MetYFV
10 9 com12 M∞MetXN∞MetYF:X1-1 ontoYxXyXxMy=FxNFyFV
11 f1oeq1 f=Ff:X1-1 ontoYF:X1-1 ontoY
12 fveq1 f=Ffx=Fx
13 fveq1 f=Ffy=Fy
14 12 13 oveq12d f=FfxNfy=FxNFy
15 14 eqeq2d f=FxMy=fxNfyxMy=FxNFy
16 15 2ralbidv f=FxXyXxMy=fxNfyxXyXxMy=FxNFy
17 11 16 anbi12d f=Ff:X1-1 ontoYxXyXxMy=fxNfyF:X1-1 ontoYxXyXxMy=FxNFy
18 17 elab3g F:X1-1 ontoYxXyXxMy=FxNFyFVFf|f:X1-1 ontoYxXyXxMy=fxNfyF:X1-1 ontoYxXyXxMy=FxNFy
19 10 18 syl M∞MetXN∞MetYFf|f:X1-1 ontoYxXyXxMy=fxNfyF:X1-1 ontoYxXyXxMy=FxNFy
20 2 19 bitrd M∞MetXN∞MetYFMIsmtyNF:X1-1 ontoYxXyXxMy=FxNFy