Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex1_GL02a_GM)
The rewrite relation of the following TRS is considered.
|
a__eq(0,0) |
→ |
true |
(1) |
|
a__eq(s(X),s(Y)) |
→ |
a__eq(X,Y) |
(2) |
|
a__eq(X,Y) |
→ |
false |
(3) |
|
a__inf(X) |
→ |
cons(X,inf(s(X))) |
(4) |
|
a__take(0,X) |
→ |
nil |
(5) |
|
a__take(s(X),cons(Y,L)) |
→ |
cons(Y,take(X,L)) |
(6) |
|
a__length(nil) |
→ |
0 |
(7) |
|
a__length(cons(X,L)) |
→ |
s(length(L)) |
(8) |
|
mark(eq(X1,X2)) |
→ |
a__eq(X1,X2) |
(9) |
|
mark(inf(X)) |
→ |
a__inf(mark(X)) |
(10) |
|
mark(take(X1,X2)) |
→ |
a__take(mark(X1),mark(X2)) |
(11) |
|
mark(length(X)) |
→ |
a__length(mark(X)) |
(12) |
|
mark(0) |
→ |
0 |
(13) |
|
mark(true) |
→ |
true |
(14) |
|
mark(s(X)) |
→ |
s(X) |
(15) |
|
mark(false) |
→ |
false |
(16) |
|
mark(cons(X1,X2)) |
→ |
cons(X1,X2) |
(17) |
|
mark(nil) |
→ |
nil |
(18) |
|
a__eq(X1,X2) |
→ |
eq(X1,X2) |
(19) |
|
a__inf(X) |
→ |
inf(X) |
(20) |
|
a__take(X1,X2) |
→ |
take(X1,X2) |
(21) |
|
a__length(X) |
→ |
length(X) |
(22) |
The evaluation strategy is innermost.Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(a__eq) |
= |
9 |
|
stat(a__eq) |
= |
mul
|
| prec(0) |
= |
1 |
|
stat(0) |
= |
mul
|
| prec(true) |
= |
2 |
|
stat(true) |
= |
mul
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
mul
|
| prec(false) |
= |
3 |
|
stat(false) |
= |
mul
|
| prec(a__inf) |
= |
4 |
|
stat(a__inf) |
= |
mul
|
| prec(cons) |
= |
0 |
|
stat(cons) |
= |
mul
|
| prec(inf) |
= |
0 |
|
stat(inf) |
= |
mul
|
| prec(a__take) |
= |
7 |
|
stat(a__take) |
= |
mul
|
| prec(nil) |
= |
5 |
|
stat(nil) |
= |
mul
|
| prec(take) |
= |
6 |
|
stat(take) |
= |
mul
|
| prec(a__length) |
= |
9 |
|
stat(a__length) |
= |
mul
|
| prec(length) |
= |
9 |
|
stat(length) |
= |
mul
|
| prec(mark) |
= |
9 |
|
stat(mark) |
= |
mul
|
| prec(eq) |
= |
8 |
|
stat(eq) |
= |
mul
|
| π(a__eq) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(true) |
= |
[] |
| π(s) |
= |
[1] |
| π(false) |
= |
[] |
| π(a__inf) |
= |
[1] |
| π(cons) |
= |
[1,2] |
| π(inf) |
= |
[1] |
| π(a__take) |
= |
[1,2] |
| π(nil) |
= |
[] |
| π(take) |
= |
[1,2] |
| π(a__length) |
= |
[1] |
| π(length) |
= |
[1] |
| π(mark) |
= |
[1] |
| π(eq) |
= |
[1,2] |
all of the following rules can be deleted.
|
a__eq(0,0) |
→ |
true |
(1) |
|
a__eq(s(X),s(Y)) |
→ |
a__eq(X,Y) |
(2) |
|
a__eq(X,Y) |
→ |
false |
(3) |
|
a__inf(X) |
→ |
cons(X,inf(s(X))) |
(4) |
|
a__take(0,X) |
→ |
nil |
(5) |
|
a__take(s(X),cons(Y,L)) |
→ |
cons(Y,take(X,L)) |
(6) |
|
a__length(nil) |
→ |
0 |
(7) |
|
a__length(cons(X,L)) |
→ |
s(length(L)) |
(8) |
|
mark(eq(X1,X2)) |
→ |
a__eq(X1,X2) |
(9) |
|
mark(inf(X)) |
→ |
a__inf(mark(X)) |
(10) |
|
mark(take(X1,X2)) |
→ |
a__take(mark(X1),mark(X2)) |
(11) |
|
mark(0) |
→ |
0 |
(13) |
|
mark(true) |
→ |
true |
(14) |
|
mark(s(X)) |
→ |
s(X) |
(15) |
|
mark(false) |
→ |
false |
(16) |
|
mark(cons(X1,X2)) |
→ |
cons(X1,X2) |
(17) |
|
mark(nil) |
→ |
nil |
(18) |
|
a__eq(X1,X2) |
→ |
eq(X1,X2) |
(19) |
|
a__inf(X) |
→ |
inf(X) |
(20) |
|
a__take(X1,X2) |
→ |
take(X1,X2) |
(21) |
1.1 Rule Removal
Using the
| prec(mark) |
= |
1 |
|
stat(mark) |
= |
lex
|
| prec(length) |
= |
0 |
|
stat(length) |
= |
lex
|
| prec(a__length) |
= |
0 |
|
stat(a__length) |
= |
lex
|
| π(mark) |
= |
[1] |
| π(length) |
= |
[1] |
| π(a__length) |
= |
[1] |
all of the following rules can be deleted.
|
mark(length(X)) |
→ |
a__length(mark(X)) |
(12) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(a__length) |
= |
1 |
|
weight(a__length) |
= |
1 |
|
|
|
| prec(length) |
= |
0 |
|
weight(length) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
a__length(X) |
→ |
length(X) |
(22) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.