Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex1_GM99_GM)
The rewrite relation of the following TRS is considered.
a__f(a,b,X) |
→ |
a__f(X,X,mark(X)) |
(1) |
a__c |
→ |
a |
(2) |
a__c |
→ |
b |
(3) |
mark(f(X1,X2,X3)) |
→ |
a__f(X1,X2,mark(X3)) |
(4) |
mark(c) |
→ |
a__c |
(5) |
mark(a) |
→ |
a |
(6) |
mark(b) |
→ |
b |
(7) |
a__f(X1,X2,X3) |
→ |
f(X1,X2,X3) |
(8) |
a__c |
→ |
c |
(9) |
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#(a,b,X) |
→ |
a__f#(X,X,mark(X)) |
(10) |
a__f#(a,b,X) |
→ |
mark#(X) |
(11) |
mark#(f(X1,X2,X3)) |
→ |
a__f#(X1,X2,mark(X3)) |
(12) |
mark#(f(X1,X2,X3)) |
→ |
mark#(X3) |
(13) |
mark#(c) |
→ |
a__c# |
(14) |
1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.