Certification Problem
Input (TPDB TRS_Innermost/AG01_innermost/#4.12a)
The rewrite relation of the following TRS is considered.
|
h(x,y) |
→ |
f(x,y,x) |
(1) |
|
f(0,1,x) |
→ |
h(x,x) |
(2) |
|
g(x,y) |
→ |
x |
(3) |
|
g(x,y) |
→ |
y |
(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.
|
h#(x,y) |
→ |
f#(x,y,x) |
(5) |
|
f#(0,1,x) |
→ |
h#(x,x) |
(6) |
1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
There are no lhss.
1.1.1.1 Instantiation Processor
We instantiate the pair
to the following set of pairs
|
h#(z0,z0) |
→ |
f#(z0,z0,z0) |
(7) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.