Certification Problem
Input (TPDB TRS_Standard/Various_04/21)
The rewrite relation of the following TRS is considered.
|
+(p1,p1) |
→ |
p2 |
(1) |
|
+(p1,+(p2,p2)) |
→ |
p5 |
(2) |
|
+(p5,p5) |
→ |
p10 |
(3) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(4) |
|
+(p1,+(p1,x)) |
→ |
+(p2,x) |
(5) |
|
+(p1,+(p2,+(p2,x))) |
→ |
+(p5,x) |
(6) |
|
+(p2,p1) |
→ |
+(p1,p2) |
(7) |
|
+(p2,+(p1,x)) |
→ |
+(p1,+(p2,x)) |
(8) |
|
+(p2,+(p2,p2)) |
→ |
+(p1,p5) |
(9) |
|
+(p2,+(p2,+(p2,x))) |
→ |
+(p1,+(p5,x)) |
(10) |
|
+(p5,p1) |
→ |
+(p1,p5) |
(11) |
|
+(p5,+(p1,x)) |
→ |
+(p1,+(p5,x)) |
(12) |
|
+(p5,p2) |
→ |
+(p2,p5) |
(13) |
|
+(p5,+(p2,x)) |
→ |
+(p2,+(p5,x)) |
(14) |
|
+(p5,+(p5,x)) |
→ |
+(p10,x) |
(15) |
|
+(p10,p1) |
→ |
+(p1,p10) |
(16) |
|
+(p10,+(p1,x)) |
→ |
+(p1,+(p10,x)) |
(17) |
|
+(p10,p2) |
→ |
+(p2,p10) |
(18) |
|
+(p10,+(p2,x)) |
→ |
+(p2,+(p10,x)) |
(19) |
|
+(p10,p5) |
→ |
+(p5,p10) |
(20) |
|
+(p10,+(p5,x)) |
→ |
+(p5,+(p10,x)) |
(21) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(p10) |
= |
2 |
|
weight(p10) |
= |
7 |
|
|
|
| prec(p5) |
= |
1 |
|
weight(p5) |
= |
5 |
|
|
|
| prec(p2) |
= |
5 |
|
weight(p2) |
= |
4 |
|
|
|
| prec(+) |
= |
3 |
|
weight(+) |
= |
1 |
|
|
|
| prec(p1) |
= |
0 |
|
weight(p1) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
+(p1,p1) |
→ |
p2 |
(1) |
|
+(p1,+(p2,p2)) |
→ |
p5 |
(2) |
|
+(p5,p5) |
→ |
p10 |
(3) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(4) |
|
+(p1,+(p1,x)) |
→ |
+(p2,x) |
(5) |
|
+(p1,+(p2,+(p2,x))) |
→ |
+(p5,x) |
(6) |
|
+(p2,p1) |
→ |
+(p1,p2) |
(7) |
|
+(p2,+(p1,x)) |
→ |
+(p1,+(p2,x)) |
(8) |
|
+(p2,+(p2,p2)) |
→ |
+(p1,p5) |
(9) |
|
+(p2,+(p2,+(p2,x))) |
→ |
+(p1,+(p5,x)) |
(10) |
|
+(p5,p1) |
→ |
+(p1,p5) |
(11) |
|
+(p5,+(p1,x)) |
→ |
+(p1,+(p5,x)) |
(12) |
|
+(p5,p2) |
→ |
+(p2,p5) |
(13) |
|
+(p5,+(p2,x)) |
→ |
+(p2,+(p5,x)) |
(14) |
|
+(p5,+(p5,x)) |
→ |
+(p10,x) |
(15) |
|
+(p10,p1) |
→ |
+(p1,p10) |
(16) |
|
+(p10,+(p1,x)) |
→ |
+(p1,+(p10,x)) |
(17) |
|
+(p10,p2) |
→ |
+(p2,p10) |
(18) |
|
+(p10,+(p2,x)) |
→ |
+(p2,+(p10,x)) |
(19) |
|
+(p10,p5) |
→ |
+(p5,p10) |
(20) |
|
+(p10,+(p5,x)) |
→ |
+(p5,+(p10,x)) |
(21) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.