Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_2_Luc02c_GM)

The rewrite relation of the following TRS is considered.

a__2nd(cons(X,cons(Y,Z))) mark(Y) (1)
a__from(X) cons(mark(X),from(s(X))) (2)
mark(2nd(X)) a__2nd(mark(X)) (3)
mark(from(X)) a__from(mark(X)) (4)
mark(cons(X1,X2)) cons(mark(X1),X2) (5)
mark(s(X)) s(mark(X)) (6)
a__2nd(X) 2nd(X) (7)
a__from(X) from(X) (8)

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__2nd#(cons(X,cons(Y,Z))) mark#(Y) (9)
a__from#(X) mark#(X) (10)
mark#(2nd(X)) a__2nd#(mark(X)) (11)
mark#(2nd(X)) mark#(X) (12)
mark#(from(X)) a__from#(mark(X)) (13)
mark#(from(X)) mark#(X) (14)
mark#(cons(X1,X2)) mark#(X1) (15)
mark#(s(X)) mark#(X) (16)

1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 1 with strict dimension 1 over the arctic semiring over the naturals
[a__2nd#(x1)] =
0
+
0
· x1
[cons(x1, x2)] =
-∞
+
1
· x1 +
0
· x2
[mark#(x1)] =
0
+
1
· x1
[a__from#(x1)] =
0
+
2
· x1
[2nd(x1)] =
1
+
0
· x1
[mark(x1)] =
-∞
+
0
· x1
[from(x1)] =
-∞
+
1
· x1
[s(x1)] =
-∞
+
0
· x1
[a__2nd(x1)] =
1
+
0
· x1
[a__from(x1)] =
-∞
+
1
· x1
the pair
mark#(2nd(X)) a__2nd#(mark(X)) (11)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.