Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/MYNAT_nosorts_FR)
The rewrite relation of the following TRS is considered.
and(tt,X) |
→ |
activate(X) |
(1) |
plus(N,0) |
→ |
N |
(2) |
plus(N,s(M)) |
→ |
s(plus(N,M)) |
(3) |
x(N,0) |
→ |
0 |
(4) |
x(N,s(M)) |
→ |
plus(x(N,M),N) |
(5) |
activate(X) |
→ |
X |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
prec(and) |
= |
1 |
|
stat(and) |
= |
mul
|
prec(tt) |
= |
2 |
|
stat(tt) |
= |
mul
|
prec(activate) |
= |
0 |
|
stat(activate) |
= |
mul
|
prec(plus) |
= |
3 |
|
stat(plus) |
= |
lex
|
prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
prec(s) |
= |
0 |
|
stat(s) |
= |
mul
|
prec(x) |
= |
4 |
|
stat(x) |
= |
mul
|
π(and) |
= |
[1,2] |
π(tt) |
= |
[] |
π(activate) |
= |
[1] |
π(plus) |
= |
[1,2] |
π(0) |
= |
[] |
π(s) |
= |
[1] |
π(x) |
= |
[1,2] |
all of the following rules can be deleted.
and(tt,X) |
→ |
activate(X) |
(1) |
plus(N,0) |
→ |
N |
(2) |
plus(N,s(M)) |
→ |
s(plus(N,M)) |
(3) |
x(N,0) |
→ |
0 |
(4) |
x(N,s(M)) |
→ |
plus(x(N,M),N) |
(5) |
activate(X) |
→ |
X |
(6) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.