Metamath Proof Explorer


Theorem funline

Description: Show that the Line relationship is a function. (Contributed by Scott Fenton, 25-Oct-2013) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion funline FunLine

Proof

Step Hyp Ref Expression
1 reeanv nma𝔼nb𝔼nabl=abColinear-1a𝔼mb𝔼mabk=abColinear-1na𝔼nb𝔼nabl=abColinear-1ma𝔼mb𝔼mabk=abColinear-1
2 eqtr3 l=abColinear-1k=abColinear-1l=k
3 2 ad2ant2l a𝔼nb𝔼nabl=abColinear-1a𝔼mb𝔼mabk=abColinear-1l=k
4 3 a1i nma𝔼nb𝔼nabl=abColinear-1a𝔼mb𝔼mabk=abColinear-1l=k
5 4 rexlimivv nma𝔼nb𝔼nabl=abColinear-1a𝔼mb𝔼mabk=abColinear-1l=k
6 1 5 sylbir na𝔼nb𝔼nabl=abColinear-1ma𝔼mb𝔼mabk=abColinear-1l=k
7 6 gen2 lkna𝔼nb𝔼nabl=abColinear-1ma𝔼mb𝔼mabk=abColinear-1l=k
8 eqeq1 l=kl=abColinear-1k=abColinear-1
9 8 anbi2d l=ka𝔼nb𝔼nabl=abColinear-1a𝔼nb𝔼nabk=abColinear-1
10 9 rexbidv l=kna𝔼nb𝔼nabl=abColinear-1na𝔼nb𝔼nabk=abColinear-1
11 fveq2 n=m𝔼n=𝔼m
12 11 eleq2d n=ma𝔼na𝔼m
13 11 eleq2d n=mb𝔼nb𝔼m
14 12 13 3anbi12d n=ma𝔼nb𝔼naba𝔼mb𝔼mab
15 14 anbi1d n=ma𝔼nb𝔼nabk=abColinear-1a𝔼mb𝔼mabk=abColinear-1
16 15 cbvrexvw na𝔼nb𝔼nabk=abColinear-1ma𝔼mb𝔼mabk=abColinear-1
17 10 16 bitrdi l=kna𝔼nb𝔼nabl=abColinear-1ma𝔼mb𝔼mabk=abColinear-1
18 17 mo4 *lna𝔼nb𝔼nabl=abColinear-1lkna𝔼nb𝔼nabl=abColinear-1ma𝔼mb𝔼mabk=abColinear-1l=k
19 7 18 mpbir *lna𝔼nb𝔼nabl=abColinear-1
20 19 funoprab Funabl|na𝔼nb𝔼nabl=abColinear-1
21 df-line2 Line=abl|na𝔼nb𝔼nabl=abColinear-1
22 21 funeqi FunLineFunabl|na𝔼nb𝔼nabl=abColinear-1
23 20 22 mpbir FunLine