Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/PEANO_nosorts_noand_GM)
The rewrite relation of the following TRS is considered.
|
a__U11(tt,M,N) |
→ |
a__U12(tt,M,N) |
(1) |
|
a__U12(tt,M,N) |
→ |
s(a__plus(mark(N),mark(M))) |
(2) |
|
a__plus(N,0) |
→ |
mark(N) |
(3) |
|
a__plus(N,s(M)) |
→ |
a__U11(tt,M,N) |
(4) |
|
mark(U11(X1,X2,X3)) |
→ |
a__U11(mark(X1),X2,X3) |
(5) |
|
mark(U12(X1,X2,X3)) |
→ |
a__U12(mark(X1),X2,X3) |
(6) |
|
mark(plus(X1,X2)) |
→ |
a__plus(mark(X1),mark(X2)) |
(7) |
|
mark(tt) |
→ |
tt |
(8) |
|
mark(s(X)) |
→ |
s(mark(X)) |
(9) |
|
mark(0) |
→ |
0 |
(10) |
|
a__U11(X1,X2,X3) |
→ |
U11(X1,X2,X3) |
(11) |
|
a__U12(X1,X2,X3) |
→ |
U12(X1,X2,X3) |
(12) |
|
a__plus(X1,X2) |
→ |
plus(X1,X2) |
(13) |
The evaluation strategy is innermost.Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(tt) |
= |
6 |
|
weight(tt) |
= |
2 |
|
|
|
| prec(0) |
= |
0 |
|
weight(0) |
= |
1 |
|
|
|
| prec(s) |
= |
1 |
|
weight(s) |
= |
2 |
|
|
|
| prec(mark) |
= |
9 |
|
weight(mark) |
= |
0 |
|
|
|
| prec(a__U11) |
= |
5 |
|
weight(a__U11) |
= |
0 |
|
|
|
| prec(a__U12) |
= |
4 |
|
weight(a__U12) |
= |
0 |
|
|
|
| prec(a__plus) |
= |
8 |
|
weight(a__plus) |
= |
0 |
|
|
|
| prec(U11) |
= |
2 |
|
weight(U11) |
= |
0 |
|
|
|
| prec(U12) |
= |
3 |
|
weight(U12) |
= |
0 |
|
|
|
| prec(plus) |
= |
7 |
|
weight(plus) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
a__U11(tt,M,N) |
→ |
a__U12(tt,M,N) |
(1) |
|
a__U12(tt,M,N) |
→ |
s(a__plus(mark(N),mark(M))) |
(2) |
|
a__plus(N,0) |
→ |
mark(N) |
(3) |
|
a__plus(N,s(M)) |
→ |
a__U11(tt,M,N) |
(4) |
|
mark(U11(X1,X2,X3)) |
→ |
a__U11(mark(X1),X2,X3) |
(5) |
|
mark(U12(X1,X2,X3)) |
→ |
a__U12(mark(X1),X2,X3) |
(6) |
|
mark(plus(X1,X2)) |
→ |
a__plus(mark(X1),mark(X2)) |
(7) |
|
mark(tt) |
→ |
tt |
(8) |
|
mark(s(X)) |
→ |
s(mark(X)) |
(9) |
|
mark(0) |
→ |
0 |
(10) |
|
a__U11(X1,X2,X3) |
→ |
U11(X1,X2,X3) |
(11) |
|
a__U12(X1,X2,X3) |
→ |
U12(X1,X2,X3) |
(12) |
|
a__plus(X1,X2) |
→ |
plus(X1,X2) |
(13) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.