let B be non empty set ; :: thesis: for A, C being set
for f being uncurrying Function
for g being currying Function st dom f c= Funcs A,(Funcs B,C) & rng f c= dom g holds
g * f = id (dom f)
let A, C be set ; :: thesis: for f being uncurrying Function
for g being currying Function st dom f c= Funcs A,(Funcs B,C) & rng f c= dom g holds
g * f = id (dom f)
let f be uncurrying Function; :: thesis: for g being currying Function st dom f c= Funcs A,(Funcs B,C) & rng f c= dom g holds
g * f = id (dom f)
let g be currying Function; :: thesis: ( dom f c= Funcs A,(Funcs B,C) & rng f c= dom g implies g * f = id (dom f) )
assume that
A1: dom f c= Funcs A,(Funcs B,C) and
A2: rng f c= dom g ; :: thesis: g * f = id (dom f)

A3: now

let x be set ; :: thesis: ( x in dom f implies (g * f) . x = x )
assume A4: x in dom f ; :: thesis: (g * f) . x = x
then reconsider X = x as Function by Def1;
A5: ex F being Function st
( X = F & dom F = A & rng F c= Funcs B,C ) by A1, A4, FUNCT_2:def 2;
A6: f . x in rng f by A4, FUNCT_1:def 5;
then reconsider Y = f . x as Function by A2, Def2;
thus (g * f) . x = g . (f . x) by A4, FUNCT_1:23
.= curry Y by A2, A6, Def2
.= curry (uncurry X) by A4, Def1
.= x by A5, FUNCT_5:55 ; :: thesis: verum

end;

dom (g * f) = dom f by A2, RELAT_1:46;
hence g * f = id (dom f) by A3, FUNCT_1:34; :: thesis: verum