Certification Problem
Input (TPDB TRS_Standard/Secret_07_TRS/secret2)
The rewrite relation of the following TRS is considered.
|
a(h,h,h,x) |
→ |
s(x) |
(1) |
|
a(l,x,s(y),h) |
→ |
a(l,x,y,s(h)) |
(2) |
|
a(l,x,s(y),s(z)) |
→ |
a(l,x,y,a(l,x,s(y),z)) |
(3) |
|
a(l,s(x),h,z) |
→ |
a(l,x,z,z) |
(4) |
|
a(s(l),h,h,z) |
→ |
a(l,z,h,z) |
(5) |
|
+(x,h) |
→ |
x |
(6) |
|
+(h,x) |
→ |
x |
(7) |
|
+(s(x),s(y)) |
→ |
s(s(+(x,y))) |
(8) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(9) |
|
s(h) |
→ |
1 |
(10) |
|
*(h,x) |
→ |
h |
(11) |
|
*(x,h) |
→ |
h |
(12) |
|
*(s(x),s(y)) |
→ |
s(+(+(*(x,y),x),y)) |
(13) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(a) |
= |
2 |
|
stat(a) |
= |
lex
|
| prec(h) |
= |
1 |
|
stat(h) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(+) |
= |
3 |
|
stat(+) |
= |
lex
|
| prec(1) |
= |
1 |
|
stat(1) |
= |
lex
|
| prec(*) |
= |
4 |
|
stat(*) |
= |
lex
|
| π(a) |
= |
[1,2,3,4] |
| π(h) |
= |
[] |
| π(s) |
= |
[1] |
| π(+) |
= |
[1,2] |
| π(1) |
= |
[] |
| π(*) |
= |
[2,1] |
all of the following rules can be deleted.
|
a(h,h,h,x) |
→ |
s(x) |
(1) |
|
a(l,x,s(y),h) |
→ |
a(l,x,y,s(h)) |
(2) |
|
a(l,x,s(y),s(z)) |
→ |
a(l,x,y,a(l,x,s(y),z)) |
(3) |
|
a(l,s(x),h,z) |
→ |
a(l,x,z,z) |
(4) |
|
a(s(l),h,h,z) |
→ |
a(l,z,h,z) |
(5) |
|
+(x,h) |
→ |
x |
(6) |
|
+(h,x) |
→ |
x |
(7) |
|
+(s(x),s(y)) |
→ |
s(s(+(x,y))) |
(8) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(9) |
|
s(h) |
→ |
1 |
(10) |
|
*(h,x) |
→ |
h |
(11) |
|
*(x,h) |
→ |
h |
(12) |
|
*(s(x),s(y)) |
→ |
s(+(+(*(x,y),x),y)) |
(13) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.