Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z065)

The rewrite relation of the following TRS is considered.

0(*(x1)) → *(1(x1)) (1)
1(*(x1)) → 0(#(x1)) (2)
#(0(x1)) → 0(#(x1)) (3)
#(1(x1)) → 1(#(x1)) (4)
#($(x1)) → *($(x1)) (5)
#(#(x1)) → #(x1) (6)
#(*(x1)) → *(x1) (7)

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
[*(x1)] = x1 +
1
[$(x1)] = x1 +
0
[#(x1)] = x1 +
1
[1(x1)] = x1 +
0
[0(x1)] = x1 +
0
all of the following rules can be deleted.
#(#(x1)) → #(x1) (6)
#(*(x1)) → *(x1) (7)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
##(1(x1)) → ##(x1) (8)
##(1(x1)) → 1#(#(x1)) (9)
##(0(x1)) → ##(x1) (10)
##(0(x1)) → 0#(#(x1)) (11)
1#(*(x1)) → ##(x1) (12)
1#(*(x1)) → 0#(#(x1)) (13)
0#(*(x1)) → 1#(x1) (14)

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
[*(x1)] = x1 +
3
[$(x1)] = x1 +
0
[#(x1)] = x1 +
3
[1(x1)] = x1 +
6
[0(x1)] = x1 +
6
[##(x1)] = x1 +
6
[1#(x1)] = x1 +
4
[0#(x1)] = x1 +
2
together with the usable rules
0(*(x1)) → *(1(x1)) (1)
1(*(x1)) → 0(#(x1)) (2)
#(0(x1)) → 0(#(x1)) (3)
#(1(x1)) → 1(#(x1)) (4)
#($(x1)) → *($(x1)) (5)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
##(1(x1)) → ##(x1) (8)
##(1(x1)) → 1#(#(x1)) (9)
##(0(x1)) → ##(x1) (10)
##(0(x1)) → 0#(#(x1)) (11)
1#(*(x1)) → ##(x1) (12)
1#(*(x1)) → 0#(#(x1)) (13)
0#(*(x1)) → 1#(x1) (14)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.