Certification Problem
Input (TPDB SRS_Standard/Trafo_06/dup09)
The rewrite relation of the following TRS is considered.
0(0(*(*(x1)))) |
→ |
*(*(1(1(x1)))) |
(1) |
1(1(*(*(x1)))) |
→ |
0(0(#(#(x1)))) |
(2) |
#(#(0(0(x1)))) |
→ |
0(0(#(#(x1)))) |
(3) |
#(#(1(1(x1)))) |
→ |
1(1(#(#(x1)))) |
(4) |
#(#($($(x1)))) |
→ |
*(*($($(x1)))) |
(5) |
#(#(#(#(x1)))) |
→ |
#(#(x1)) |
(6) |
#(#(*(*(x1)))) |
→ |
*(*(x1)) |
(7) |
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($) |
= |
0 |
|
weight($) |
= |
2 |
|
|
|
prec(#) |
= |
5 |
|
weight(#) |
= |
1 |
|
|
|
prec(1) |
= |
4 |
|
weight(1) |
= |
1 |
|
|
|
prec(0) |
= |
2 |
|
weight(0) |
= |
1 |
|
|
|
prec(*) |
= |
1 |
|
weight(*) |
= |
1 |
|
|
|
all of the following rules can be deleted.
0(0(*(*(x1)))) |
→ |
*(*(1(1(x1)))) |
(1) |
1(1(*(*(x1)))) |
→ |
0(0(#(#(x1)))) |
(2) |
#(#(0(0(x1)))) |
→ |
0(0(#(#(x1)))) |
(3) |
#(#(1(1(x1)))) |
→ |
1(1(#(#(x1)))) |
(4) |
#(#($($(x1)))) |
→ |
*(*($($(x1)))) |
(5) |
#(#(#(#(x1)))) |
→ |
#(#(x1)) |
(6) |
#(#(*(*(x1)))) |
→ |
*(*(x1)) |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.