Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex4_4_Luc96b_GM)
The rewrite relation of the following TRS is considered.
|
a__f(g(X),Y) |
→ |
a__f(mark(X),f(g(X),Y)) |
(1) |
|
mark(f(X1,X2)) |
→ |
a__f(mark(X1),X2) |
(2) |
|
mark(g(X)) |
→ |
g(mark(X)) |
(3) |
|
a__f(X1,X2) |
→ |
f(X1,X2) |
(4) |
The evaluation strategy is innermost.Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a__f#(g(X),Y) |
→ |
a__f#(mark(X),f(g(X),Y)) |
(5) |
|
a__f#(g(X),Y) |
→ |
mark#(X) |
(6) |
|
mark#(f(X1,X2)) |
→ |
a__f#(mark(X1),X2) |
(7) |
|
mark#(f(X1,X2)) |
→ |
mark#(X1) |
(8) |
|
mark#(g(X)) |
→ |
mark#(X) |
(9) |
1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [a__f#(x1, x2)] |
= |
2 + x1
|
| [mark(x1)] |
= |
1 + 2 · x1
|
| [f(x1, x2)] |
= |
2 + 2 · x1
|
| [a__f(x1, x2)] |
= |
2 + 2 · x1
|
| [g(x1)] |
= |
1 + 2 · x1
|
| [mark#(x1)] |
= |
2 + x1
|
the
pairs
|
a__f#(g(X),Y) |
→ |
mark#(X) |
(6) |
|
mark#(f(X1,X2)) |
→ |
a__f#(mark(X1),X2) |
(7) |
|
mark#(f(X1,X2)) |
→ |
mark#(X1) |
(8) |
|
mark#(g(X)) |
→ |
mark#(X) |
(9) |
could be deleted.
1.1.1 Reduction Pair Processor
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(g) |
= |
2 |
|
weight(g) |
= |
2 |
|
|
|
| prec(mark) |
= |
3 |
|
weight(mark) |
= |
1 |
|
|
|
| prec(f) |
= |
0 |
|
weight(f) |
= |
1 |
|
|
|
| prec(a__f) |
= |
1 |
|
weight(a__f) |
= |
1 |
|
|
|
in combination with the following argument filter
| π(a__f#) |
= |
1 |
| π(g) |
= |
[1] |
| π(mark) |
= |
[1] |
| π(f) |
= |
[] |
| π(a__f) |
= |
[] |
the
pair
|
a__f#(g(X),Y) |
→ |
a__f#(mark(X),f(g(X),Y)) |
(5) |
could be deleted.
1.1.1.1 P is empty
There are no pairs anymore.