L¨ osungsvorschlag Termersetzungssysteme – Blatt 1, Aufgabe 1
Wir benutzen hier die Schreibweise x+y f¨urplus(x, y) und x∗y f¨ur times(x, y).
a)
D(X) → 1 D(Y) → 0 D(1) → 0 D(0) → 0
D(p+q) → D(p) +D(q)
D(p∗q) → (D(p)∗q) + (p∗D(q)) b)
D(X∗1)
(D(X)∗1) + (X∗D(1))
ttiiiiiiiiiiiiiiiii
**
UU UU UU UU UU UU UU UU U
(1∗1) + (X∗D(1))
**
UU UU UU UU UU UU UU UU U
(D(X)∗1) + (X∗0)
ttiiiiiiiiiiiiiiiii
(1∗1) + (X∗0) Das Ergebnis ist eindeutig.
c)
D(X∗1)
vvnnnnnnnnnnnnn
**
UU UU UU UU UU UU UU UU UU
D(X)
(D(X)∗1) + (X∗D(1))
ttiiiiiiiiiiiiiiiii
**UUUUUUUUUUUUUUUUU
1 (1∗1) + (X∗D(1))
**
UU UU UU UU UU UU UU UU U
(D(X)∗1) + (X∗0)
**UUUUUUUUUUUUUUUUU
D(X) + (X∗D(1))
qqdddddddddddddddddddddddddddddddddddddd
1+ (X∗D(1))
**
UU UU UU UU UU UU UU UU U
(1∗1) + (X∗0)
D(X) + (X∗0)
ttiiiiiiiiiiiiiiiii
1+ (X∗0) Das Ergebnis ist nicht eindeutig.