Metamath Proof Explorer


Theorem dffrege115

Description: If from the circumstance that c is a result of an application of the procedure R to b , whatever b may be, it can be inferred that every result of an application of the procedure R to b is the same as c , then we say : "The procedure R is single-valued". Definition 115 of Frege1879 p. 77. (Contributed by RP, 7-Jul-2020)

Ref Expression
Assertion dffrege115 c b b R c a b R a a = c Fun R -1 -1

Proof

Step Hyp Ref Expression
1 alcom c b b R c a b R a a = c b c b R c a b R a a = c
2 19.21v a b R c b R a a = c b R c a b R a a = c
3 impexp b R c b R a a = c b R c b R a a = c
4 vex b V
5 vex c V
6 4 5 brcnv b R -1 -1 c c R -1 b
7 df-br b R -1 -1 c b c R -1 -1
8 5 4 brcnv c R -1 b b R c
9 6 7 8 3bitr3ri b R c b c R -1 -1
10 vex a V
11 4 10 brcnv b R -1 -1 a a R -1 b
12 df-br b R -1 -1 a b a R -1 -1
13 10 4 brcnv a R -1 b b R a
14 11 12 13 3bitr3ri b R a b a R -1 -1
15 9 14 anbi12ci b R c b R a b a R -1 -1 b c R -1 -1
16 15 imbi1i b R c b R a a = c b a R -1 -1 b c R -1 -1 a = c
17 3 16 bitr3i b R c b R a a = c b a R -1 -1 b c R -1 -1 a = c
18 17 albii a b R c b R a a = c a b a R -1 -1 b c R -1 -1 a = c
19 2 18 bitr3i b R c a b R a a = c a b a R -1 -1 b c R -1 -1 a = c
20 19 albii c b R c a b R a a = c c a b a R -1 -1 b c R -1 -1 a = c
21 alcom c a b a R -1 -1 b c R -1 -1 a = c a c b a R -1 -1 b c R -1 -1 a = c
22 20 21 bitri c b R c a b R a a = c a c b a R -1 -1 b c R -1 -1 a = c
23 opeq2 a = c b a = b c
24 23 eleq1d a = c b a R -1 -1 b c R -1 -1
25 24 mo4 * a b a R -1 -1 a c b a R -1 -1 b c R -1 -1 a = c
26 df-mo * a b a R -1 -1 c a b a R -1 -1 a = c
27 22 25 26 3bitr2i c b R c a b R a a = c c a b a R -1 -1 a = c
28 27 albii b c b R c a b R a a = c b c a b a R -1 -1 a = c
29 relcnv Rel R -1 -1
30 29 biantrur b c a b a R -1 -1 a = c Rel R -1 -1 b c a b a R -1 -1 a = c
31 dffun5 Fun R -1 -1 Rel R -1 -1 b c a b a R -1 -1 a = c
32 30 31 bitr4i b c a b a R -1 -1 a = c Fun R -1 -1
33 1 28 32 3bitri c b b R c a b R a a = c Fun R -1 -1