Certification Problem
Input (TPDB TRS_Standard/SK90/4.43)
The rewrite relation of the following TRS is considered.
|
+(x,0) |
→ |
x |
(1) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(4) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(5) |
|
f(g(f(x))) |
→ |
f(h(s(0),x)) |
(6) |
|
f(g(h(x,y))) |
→ |
f(h(s(x),y)) |
(7) |
|
f(h(x,h(y,z))) |
→ |
f(h(+(x,y),z)) |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [+(x1, x2)] |
= |
1 + 1 · x1 + 2 · x2
|
| [0] |
= |
0 |
| [s(x1)] |
= |
1 · x1
|
| [f(x1)] |
= |
2 + 1 · x1
|
| [g(x1)] |
= |
1 + 2 · x1
|
| [h(x1, x2)] |
= |
2 + 1 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
+(x,0) |
→ |
x |
(1) |
|
+(0,y) |
→ |
y |
(3) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(5) |
|
f(g(f(x))) |
→ |
f(h(s(0),x)) |
(6) |
|
f(g(h(x,y))) |
→ |
f(h(s(x),y)) |
(7) |
|
f(h(x,h(y,z))) |
→ |
f(h(+(x,y),z)) |
(8) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(s) |
= |
0 |
|
weight(s) |
= |
1 |
|
|
|
| prec(+) |
= |
1 |
|
weight(+) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(2) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(4) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.