Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/247906)

The rewrite relation of the following TRS is considered.

0(x1) → 1(x1) (1)
0(0(x1)) → 0(x1) (2)
3(4(5(x1))) → 4(3(5(x1))) (3)
2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) → 0(0(1(0(1(1(0(1(1(1(1(1(1(1(1(0(0(0(0(0(0(1(1(1(1(1(0(0(1(1(1(1(0(0(1(1(1(1(0(1(1(0(0(0(0(0(1(1(0(0(0(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) (4)
1(1(1(1(1(1(0(1(1(1(1(1(1(0(0(0(1(0(1(1(1(0(1(1(1(0(1(0(1(0(1(1(0(1(0(0(1(1(0(1(0(1(0(1(0(0(0(0(0(0(0(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) → 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
0
[4(x1)] = x1 +
0
[3(x1)] = x1 +
0
[2(x1)] = x1 +
28815/103
[1(x1)] = x1 +
8730/103
[0(x1)] = x1 +
28712/103
all of the following rules can be deleted.
0(x1) → 1(x1) (1)
0(0(x1)) → 0(x1) (2)
2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) → 0(0(1(0(1(1(0(1(1(1(1(1(1(1(1(0(0(0(0(0(0(1(1(1(1(1(0(0(1(1(1(1(0(0(1(1(1(1(0(1(1(0(0(0(0(0(1(1(0(0(0(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) (4)
1(1(1(1(1(1(0(1(1(1(1(1(1(0(0(0(1(0(1(1(1(0(1(1(1(0(1(0(1(0(1(1(0(1(0(0(1(1(0(1(0(1(0(1(0(0(0(0(0(0(0(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) → 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) (5)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
3#(4(5(x1))) → 3#(5(x1)) (6)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
0
[4(x1)] = x1 +
1
[3(x1)] = x1 +
0
[3#(x1)] = x1 +
0
together with the usable rule
3(4(5(x1))) → 4(3(5(x1))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
3#(4(5(x1))) → 3#(5(x1)) (6)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.