Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/157388)

The rewrite relation of the following TRS is considered.

There are 151 ruless (increase limit for explicit display).

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

There are 151 ruless (increase limit for explicit display).

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 199 ruless (increase limit for explicit display).

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 +
0
[3(x1)] = x1 +
0
[2(x1)] = x1 +
0
[1(x1)] = x1 +
0
[0(x1)] = x1 +
1
[1#(x1)] = x1 +
0
together with the usable rules

There are 151 ruless (increase limit for explicit display).

(w.r.t. the implicit argument filter of the reduction pair), the pairs
1#(1(5(4(0(x1))))) 1#(2(5(x1))) (316)
1#(1(3(4(0(x1))))) 1#(4(1(2(x1)))) (345)
1#(1(3(4(0(x1))))) 1#(2(x1)) (347)
1#(1(3(0(0(x1))))) 1#(3(1(2(x1)))) (349)
1#(1(3(0(0(x1))))) 1#(2(x1)) (350)
1#(1(0(3(4(x1))))) 1#(x1) (363)
1#(1(0(3(4(x1))))) 1#(3(x1)) (365)
1#(1(0(3(4(x1))))) 1#(3(4(x1))) (366)
1#(1(0(3(4(x1))))) 1#(3(1(x1))) (367)
1#(1(0(3(4(x1))))) 1#(1(3(4(x1)))) (369)
1#(1(0(1(4(x1))))) 1#(x1) (370)
1#(0(5(3(4(x1))))) 1#(x1) (373)
1#(0(5(3(4(x1))))) 1#(5(x1)) (374)
1#(0(5(3(4(x1))))) 1#(5(3(x1))) (375)
1#(0(5(3(4(x1))))) 1#(4(3(x1))) (376)
1#(0(5(3(4(x1))))) 1#(3(x1)) (377)
1#(0(5(1(4(x1))))) 1#(x1) (381)
1#(0(5(1(4(x1))))) 1#(1(x1)) (382)
1#(0(3(5(4(x1))))) 1#(x1) (385)
1#(0(3(5(4(x1))))) 1#(5(x1)) (386)
1#(0(3(5(4(x1))))) 1#(4(x1)) (389)
1#(0(3(5(4(x1))))) 1#(4(5(x1))) (390)
1#(0(3(5(4(x1))))) 1#(4(2(x1))) (393)
1#(0(3(5(4(x1))))) 1#(4(2(5(x1)))) (394)
1#(0(3(5(4(x1))))) 1#(2(x1)) (401)
1#(0(3(5(4(x1))))) 1#(2(5(x1))) (402)
1#(0(3(4(x1)))) 1#(x1) (410)
1#(0(3(4(x1)))) 1#(4(3(x1))) (411)
1#(0(3(4(x1)))) 1#(3(x1)) (414)
1#(0(3(4(x1)))) 1#(3(4(x1))) (415)
1#(0(3(4(0(x1))))) 1#(3(2(x1))) (441)
1#(0(2(1(4(x1))))) 1#(x1) (454)
1#(0(1(5(4(x1))))) 1#(x1) (461)
1#(0(1(5(4(x1))))) 1#(5(x1)) (462)
1#(0(1(5(4(x1))))) 1#(4(5(x1))) (464)
1#(0(1(4(x1)))) 1#(x1) (474)
1#(0(1(4(0(x1))))) 1#(2(x1)) (488)
1#(0(1(3(4(x1))))) 1#(x1) (490)
1#(0(1(0(x1)))) 1#(3(1(2(x1)))) (496)
1#(0(1(0(x1)))) 1#(2(x1)) (498)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.