YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
B(x0) |
→ |
W(M(M(V(x0)))) |
M(x0) |
→ |
x0 |
M(V(a(x0))) |
→ |
V(Xa(x0)) |
M(V(b(x0))) |
→ |
V(Xb(x0)) |
M(V(c(x0))) |
→ |
V(Xc(x0)) |
M(V(d(x0))) |
→ |
V(Xd(x0)) |
Xa(a(x0)) |
→ |
a(Xa(x0)) |
Xa(b(x0)) |
→ |
b(Xa(x0)) |
Xa(c(x0)) |
→ |
c(Xa(x0)) |
Xa(d(x0)) |
→ |
d(Xa(x0)) |
Xb(a(x0)) |
→ |
a(Xb(x0)) |
Xb(b(x0)) |
→ |
b(Xb(x0)) |
Xb(c(x0)) |
→ |
c(Xb(x0)) |
Xb(d(x0)) |
→ |
d(Xb(x0)) |
Xc(a(x0)) |
→ |
a(Xc(x0)) |
Xc(b(x0)) |
→ |
b(Xc(x0)) |
Xc(c(x0)) |
→ |
c(Xc(x0)) |
Xc(d(x0)) |
→ |
d(Xc(x0)) |
Xd(a(x0)) |
→ |
a(Xd(x0)) |
Xd(b(x0)) |
→ |
b(Xd(x0)) |
Xd(c(x0)) |
→ |
c(Xd(x0)) |
Xd(d(x0)) |
→ |
d(Xd(x0)) |
Xa(E(x0)) |
→ |
a(E(x0)) |
Xb(E(x0)) |
→ |
b(E(x0)) |
Xc(E(x0)) |
→ |
c(E(x0)) |
Xd(E(x0)) |
→ |
d(E(x0)) |
W(V(x0)) |
→ |
R(L(x0)) |
L(a(x0)) |
→ |
Ya(L(x0)) |
L(b(x0)) |
→ |
Yb(L(x0)) |
L(c(x0)) |
→ |
Yc(L(x0)) |
L(d(x0)) |
→ |
Yd(L(x0)) |
L(a(a(a(x0)))) |
→ |
D(b(b(x0))) |
L(b(b(b(x0)))) |
→ |
D(c(d(x0))) |
L(c(x0)) |
→ |
D(a(a(x0))) |
L(d(x0)) |
→ |
D(c(x0)) |
Ya(D(x0)) |
→ |
D(a(x0)) |
Yb(D(x0)) |
→ |
D(b(x0)) |
Yc(D(x0)) |
→ |
D(c(x0)) |
Yd(D(x0)) |
→ |
D(d(x0)) |
R(D(x0)) |
→ |
B(x0) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
[Yd(x1)] |
= |
4 ·
x1 +
-∞
|
[a(x1)] |
= |
2 ·
x1 +
-∞
|
[Ya(x1)] |
= |
2 ·
x1 +
-∞
|
[c(x1)] |
= |
4 ·
x1 +
-∞
|
[V(x1)] |
= |
0 ·
x1 +
-∞
|
[W(x1)] |
= |
2 ·
x1 +
-∞
|
[Yb(x1)] |
= |
3 ·
x1 +
-∞
|
[D(x1)] |
= |
2 ·
x1 +
-∞
|
[Yc(x1)] |
= |
4 ·
x1 +
-∞
|
[Xa(x1)] |
= |
2 ·
x1 +
-∞
|
[d(x1)] |
= |
4 ·
x1 +
-∞
|
[B(x1)] |
= |
2 ·
x1 +
-∞
|
[Xb(x1)] |
= |
3 ·
x1 +
-∞
|
[R(x1)] |
= |
0 ·
x1 +
-∞
|
[E(x1)] |
= |
0 ·
x1 +
-∞
|
[Xd(x1)] |
= |
4 ·
x1 +
-∞
|
[Xc(x1)] |
= |
4 ·
x1 +
-∞
|
[L(x1)] |
= |
2 ·
x1 +
-∞
|
[b(x1)] |
= |
3 ·
x1 +
-∞
|
[M(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
B(x0) |
→ |
W(M(M(V(x0)))) |
M(x0) |
→ |
x0 |
M(V(a(x0))) |
→ |
V(Xa(x0)) |
M(V(b(x0))) |
→ |
V(Xb(x0)) |
M(V(c(x0))) |
→ |
V(Xc(x0)) |
M(V(d(x0))) |
→ |
V(Xd(x0)) |
Xa(a(x0)) |
→ |
a(Xa(x0)) |
Xa(b(x0)) |
→ |
b(Xa(x0)) |
Xa(c(x0)) |
→ |
c(Xa(x0)) |
Xa(d(x0)) |
→ |
d(Xa(x0)) |
Xb(a(x0)) |
→ |
a(Xb(x0)) |
Xb(b(x0)) |
→ |
b(Xb(x0)) |
Xb(c(x0)) |
→ |
c(Xb(x0)) |
Xb(d(x0)) |
→ |
d(Xb(x0)) |
Xc(a(x0)) |
→ |
a(Xc(x0)) |
Xc(b(x0)) |
→ |
b(Xc(x0)) |
Xc(c(x0)) |
→ |
c(Xc(x0)) |
Xc(d(x0)) |
→ |
d(Xc(x0)) |
Xd(a(x0)) |
→ |
a(Xd(x0)) |
Xd(b(x0)) |
→ |
b(Xd(x0)) |
Xd(c(x0)) |
→ |
c(Xd(x0)) |
Xd(d(x0)) |
→ |
d(Xd(x0)) |
Xa(E(x0)) |
→ |
a(E(x0)) |
Xb(E(x0)) |
→ |
b(E(x0)) |
Xc(E(x0)) |
→ |
c(E(x0)) |
Xd(E(x0)) |
→ |
d(E(x0)) |
W(V(x0)) |
→ |
R(L(x0)) |
L(a(x0)) |
→ |
Ya(L(x0)) |
L(b(x0)) |
→ |
Yb(L(x0)) |
L(c(x0)) |
→ |
Yc(L(x0)) |
L(d(x0)) |
→ |
Yd(L(x0)) |
L(a(a(a(x0)))) |
→ |
D(b(b(x0))) |
L(c(x0)) |
→ |
D(a(a(x0))) |
L(d(x0)) |
→ |
D(c(x0)) |
Ya(D(x0)) |
→ |
D(a(x0)) |
Yb(D(x0)) |
→ |
D(b(x0)) |
Yc(D(x0)) |
→ |
D(c(x0)) |
Yd(D(x0)) |
→ |
D(d(x0)) |
R(D(x0)) |
→ |
B(x0) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
[Yd(x1)] |
= |
1 ·
x1 +
-∞
|
[a(x1)] |
= |
0 ·
x1 +
-∞
|
[Ya(x1)] |
= |
0 ·
x1 +
-∞
|
[c(x1)] |
= |
0 ·
x1 +
-∞
|
[V(x1)] |
= |
0 ·
x1 +
-∞
|
[W(x1)] |
= |
0 ·
x1 +
-∞
|
[Yb(x1)] |
= |
0 ·
x1 +
-∞
|
[D(x1)] |
= |
0 ·
x1 +
-∞
|
[Yc(x1)] |
= |
0 ·
x1 +
-∞
|
[Xa(x1)] |
= |
0 ·
x1 +
-∞
|
[d(x1)] |
= |
1 ·
x1 +
-∞
|
[B(x1)] |
= |
0 ·
x1 +
-∞
|
[Xb(x1)] |
= |
0 ·
x1 +
-∞
|
[R(x1)] |
= |
0 ·
x1 +
-∞
|
[E(x1)] |
= |
0 ·
x1 +
-∞
|
[Xd(x1)] |
= |
1 ·
x1 +
-∞
|
[Xc(x1)] |
= |
0 ·
x1 +
-∞
|
[L(x1)] |
= |
0 ·
x1 +
-∞
|
[b(x1)] |
= |
0 ·
x1 +
-∞
|
[M(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
B(x0) |
→ |
W(M(M(V(x0)))) |
M(x0) |
→ |
x0 |
M(V(a(x0))) |
→ |
V(Xa(x0)) |
M(V(b(x0))) |
→ |
V(Xb(x0)) |
M(V(c(x0))) |
→ |
V(Xc(x0)) |
M(V(d(x0))) |
→ |
V(Xd(x0)) |
Xa(a(x0)) |
→ |
a(Xa(x0)) |
Xa(b(x0)) |
→ |
b(Xa(x0)) |
Xa(c(x0)) |
→ |
c(Xa(x0)) |
Xa(d(x0)) |
→ |
d(Xa(x0)) |
Xb(a(x0)) |
→ |
a(Xb(x0)) |
Xb(b(x0)) |
→ |
b(Xb(x0)) |
Xb(c(x0)) |
→ |
c(Xb(x0)) |
Xb(d(x0)) |
→ |
d(Xb(x0)) |
Xc(a(x0)) |
→ |
a(Xc(x0)) |
Xc(b(x0)) |
→ |
b(Xc(x0)) |
Xc(c(x0)) |
→ |
c(Xc(x0)) |
Xc(d(x0)) |
→ |
d(Xc(x0)) |
Xd(a(x0)) |
→ |
a(Xd(x0)) |
Xd(b(x0)) |
→ |
b(Xd(x0)) |
Xd(c(x0)) |
→ |
c(Xd(x0)) |
Xd(d(x0)) |
→ |
d(Xd(x0)) |
Xa(E(x0)) |
→ |
a(E(x0)) |
Xb(E(x0)) |
→ |
b(E(x0)) |
Xc(E(x0)) |
→ |
c(E(x0)) |
Xd(E(x0)) |
→ |
d(E(x0)) |
W(V(x0)) |
→ |
R(L(x0)) |
L(a(x0)) |
→ |
Ya(L(x0)) |
L(b(x0)) |
→ |
Yb(L(x0)) |
L(c(x0)) |
→ |
Yc(L(x0)) |
L(d(x0)) |
→ |
Yd(L(x0)) |
L(a(a(a(x0)))) |
→ |
D(b(b(x0))) |
L(c(x0)) |
→ |
D(a(a(x0))) |
Ya(D(x0)) |
→ |
D(a(x0)) |
Yb(D(x0)) |
→ |
D(b(x0)) |
Yc(D(x0)) |
→ |
D(c(x0)) |
Yd(D(x0)) |
→ |
D(d(x0)) |
R(D(x0)) |
→ |
B(x0) |
remain.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
[Yd(x1)] |
= |
1 ·
x1 +
-∞
|
[a(x1)] |
= |
2 ·
x1 +
-∞
|
[Ya(x1)] |
= |
2 ·
x1 +
-∞
|
[c(x1)] |
= |
8 ·
x1 +
-∞
|
[V(x1)] |
= |
3 ·
x1 +
-∞
|
[W(x1)] |
= |
1 ·
x1 +
-∞
|
[Yb(x1)] |
= |
2 ·
x1 +
-∞
|
[D(x1)] |
= |
2 ·
x1 +
-∞
|
[Yc(x1)] |
= |
8 ·
x1 +
-∞
|
[Xa(x1)] |
= |
2 ·
x1 +
-∞
|
[d(x1)] |
= |
1 ·
x1 +
-∞
|
[B(x1)] |
= |
5 ·
x1 +
-∞
|
[Xb(x1)] |
= |
2 ·
x1 +
-∞
|
[R(x1)] |
= |
4 ·
x1 +
-∞
|
[E(x1)] |
= |
0 ·
x1 +
-∞
|
[Xd(x1)] |
= |
1 ·
x1 +
-∞
|
[Xc(x1)] |
= |
8 ·
x1 +
-∞
|
[L(x1)] |
= |
0 ·
x1 +
-∞
|
[b(x1)] |
= |
2 ·
x1 +
-∞
|
[M(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
M(x0) |
→ |
x0 |
M(V(a(x0))) |
→ |
V(Xa(x0)) |
M(V(b(x0))) |
→ |
V(Xb(x0)) |
M(V(c(x0))) |
→ |
V(Xc(x0)) |
M(V(d(x0))) |
→ |
V(Xd(x0)) |
Xa(a(x0)) |
→ |
a(Xa(x0)) |
Xa(b(x0)) |
→ |
b(Xa(x0)) |
Xa(c(x0)) |
→ |
c(Xa(x0)) |
Xa(d(x0)) |
→ |
d(Xa(x0)) |
Xb(a(x0)) |
→ |
a(Xb(x0)) |
Xb(b(x0)) |
→ |
b(Xb(x0)) |
Xb(c(x0)) |
→ |
c(Xb(x0)) |
Xb(d(x0)) |
→ |
d(Xb(x0)) |
Xc(a(x0)) |
→ |
a(Xc(x0)) |
Xc(b(x0)) |
→ |
b(Xc(x0)) |
Xc(c(x0)) |
→ |
c(Xc(x0)) |
Xc(d(x0)) |
→ |
d(Xc(x0)) |
Xd(a(x0)) |
→ |
a(Xd(x0)) |
Xd(b(x0)) |
→ |
b(Xd(x0)) |
Xd(c(x0)) |
→ |
c(Xd(x0)) |
Xd(d(x0)) |
→ |
d(Xd(x0)) |
Xa(E(x0)) |
→ |
a(E(x0)) |
Xb(E(x0)) |
→ |
b(E(x0)) |
Xc(E(x0)) |
→ |
c(E(x0)) |
Xd(E(x0)) |
→ |
d(E(x0)) |
W(V(x0)) |
→ |
R(L(x0)) |
L(a(x0)) |
→ |
Ya(L(x0)) |
L(b(x0)) |
→ |
Yb(L(x0)) |
L(c(x0)) |
→ |
Yc(L(x0)) |
L(d(x0)) |
→ |
Yd(L(x0)) |
L(a(a(a(x0)))) |
→ |
D(b(b(x0))) |
Ya(D(x0)) |
→ |
D(a(x0)) |
Yb(D(x0)) |
→ |
D(b(x0)) |
Yc(D(x0)) |
→ |
D(c(x0)) |
Yd(D(x0)) |
→ |
D(d(x0)) |
remain.
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
prec(D) |
= |
0 |
|
weight(D) |
= |
1 |
|
|
|
prec(Yd) |
= |
2 |
|
weight(Yd) |
= |
1 |
|
|
|
prec(Yc) |
= |
2 |
|
weight(Yc) |
= |
1 |
|
|
|
prec(Yb) |
= |
2 |
|
weight(Yb) |
= |
1 |
|
|
|
prec(Ya) |
= |
2 |
|
weight(Ya) |
= |
1 |
|
|
|
prec(R) |
= |
0 |
|
weight(R) |
= |
1 |
|
|
|
prec(L) |
= |
3 |
|
weight(L) |
= |
1 |
|
|
|
prec(E) |
= |
0 |
|
weight(E) |
= |
1 |
|
|
|
prec(Xd) |
= |
1 |
|
weight(Xd) |
= |
1 |
|
|
|
prec(d) |
= |
0 |
|
weight(d) |
= |
1 |
|
|
|
prec(Xc) |
= |
5 |
|
weight(Xc) |
= |
1 |
|
|
|
prec(c) |
= |
0 |
|
weight(c) |
= |
1 |
|
|
|
prec(Xb) |
= |
6 |
|
weight(Xb) |
= |
1 |
|
|
|
prec(b) |
= |
0 |
|
weight(b) |
= |
1 |
|
|
|
prec(Xa) |
= |
6 |
|
weight(Xa) |
= |
1 |
|
|
|
prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
prec(W) |
= |
1 |
|
weight(W) |
= |
1 |
|
|
|
prec(M) |
= |
0 |
|
weight(M) |
= |
1 |
|
|
|
prec(V) |
= |
7 |
|
weight(V) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.