YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
R(E(x0)) |
→ |
L(E(x0)) |
|
a(L(x0)) |
→ |
L(Aa(x0)) |
|
b(L(x0)) |
→ |
L(Ab(x0)) |
|
c(L(x0)) |
→ |
L(Ac(x0)) |
|
R(Aa(x0)) |
→ |
a(R(x0)) |
|
R(Ab(x0)) |
→ |
b(R(x0)) |
|
R(Ac(x0)) |
→ |
c(R(x0)) |
|
a(b(L(x0))) |
→ |
b(a(a(R(x0)))) |
|
c(b(L(x0))) |
→ |
b(b(c(R(x0)))) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [Aa(x1)] |
= |
1 ·
x1 + 0 |
| [Ac(x1)] |
= |
3 ·
x1 + 0 |
| [R(x1)] |
= |
8 ·
x1 + 0 |
| [a(x1)] |
= |
1 ·
x1 + 0 |
| [b(x1)] |
= |
1 ·
x1 + 8 |
| [E(x1)] |
= |
2 ·
x1 + 0 |
| [c(x1)] |
= |
3 ·
x1 + 0 |
| [Ab(x1)] |
= |
1 ·
x1 + 1 |
| [L(x1)] |
= |
8 ·
x1 + 0 |
the
rules
|
R(E(x0)) |
→ |
L(E(x0)) |
|
a(L(x0)) |
→ |
L(Aa(x0)) |
|
b(L(x0)) |
→ |
L(Ab(x0)) |
|
c(L(x0)) |
→ |
L(Ac(x0)) |
|
R(Aa(x0)) |
→ |
a(R(x0)) |
|
R(Ab(x0)) |
→ |
b(R(x0)) |
|
R(Ac(x0)) |
→ |
c(R(x0)) |
|
a(b(L(x0))) |
→ |
b(a(a(R(x0)))) |
remain.
1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
E(R(x0)) |
→ |
E(L(x0)) |
|
L(a(x0)) |
→ |
Aa(L(x0)) |
|
L(b(x0)) |
→ |
Ab(L(x0)) |
|
L(c(x0)) |
→ |
Ac(L(x0)) |
|
Aa(R(x0)) |
→ |
R(a(x0)) |
|
Ab(R(x0)) |
→ |
R(b(x0)) |
|
Ac(R(x0)) |
→ |
R(c(x0)) |
|
L(b(a(x0))) |
→ |
R(a(a(b(x0)))) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [Aa(x1)] |
= |
·
x1 +
|
| [Ac(x1)] |
= |
·
x1 +
|
| [R(x1)] |
= |
·
x1 +
|
| [a(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [E(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
| [Ab(x1)] |
= |
·
x1 +
|
| [L(x1)] |
= |
·
x1 +
|
the
rules
|
L(a(x0)) |
→ |
Aa(L(x0)) |
|
L(b(x0)) |
→ |
Ab(L(x0)) |
|
L(c(x0)) |
→ |
Ac(L(x0)) |
|
Aa(R(x0)) |
→ |
R(a(x0)) |
|
Ab(R(x0)) |
→ |
R(b(x0)) |
|
Ac(R(x0)) |
→ |
R(c(x0)) |
|
L(b(a(x0))) |
→ |
R(a(a(b(x0)))) |
remain.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [Aa(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ac(x1)] |
= |
7 ·
x1 +
-∞
|
| [R(x1)] |
= |
0 ·
x1 +
-∞
|
| [a(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
4 ·
x1 +
-∞
|
| [c(x1)] |
= |
7 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
4 ·
x1 +
-∞
|
| [L(x1)] |
= |
1 ·
x1 +
-∞
|
the
rules
|
L(a(x0)) |
→ |
Aa(L(x0)) |
|
L(b(x0)) |
→ |
Ab(L(x0)) |
|
L(c(x0)) |
→ |
Ac(L(x0)) |
|
Aa(R(x0)) |
→ |
R(a(x0)) |
|
Ab(R(x0)) |
→ |
R(b(x0)) |
|
Ac(R(x0)) |
→ |
R(c(x0)) |
remain.
1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(Ac) |
= |
2 |
|
weight(Ac) |
= |
1 |
|
|
|
| prec(c) |
= |
0 |
|
weight(c) |
= |
1 |
|
|
|
| prec(Ab) |
= |
2 |
|
weight(Ab) |
= |
1 |
|
|
|
| prec(b) |
= |
0 |
|
weight(b) |
= |
1 |
|
|
|
| prec(Aa) |
= |
2 |
|
weight(Aa) |
= |
1 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
| prec(L) |
= |
3 |
|
weight(L) |
= |
0 |
|
|
|
| prec(R) |
= |
0 |
|
weight(R) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.