Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup09)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
*(*(0(0(x1)))) 1(1(*(*(x1)))) (8)
*(*(1(1(x1)))) #(#(0(0(x1)))) (9)
0(0(#(#(x1)))) #(#(0(0(x1)))) (10)
1(1(#(#(x1)))) #(#(1(1(x1)))) (11)
$($(#(#(x1)))) $($(*(*(x1)))) (12)
#(#(#(#(x1)))) #(#(x1)) (6)
*(*(#(#(x1)))) *(*(x1)) (13)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[*(x1)] = 1 · x1 + 1
[0(x1)] = 1 · x1
[1(x1)] = 1 · x1
[#(x1)] = 1 · x1 + 1
[$(x1)] = 1 · x1
all of the following rules can be deleted.
#(#(#(#(x1)))) #(#(x1)) (6)
*(*(#(#(x1)))) *(*(x1)) (13)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
*#(*(0(0(x1)))) 1#(1(*(*(x1)))) (14)
*#(*(0(0(x1)))) 1#(*(*(x1))) (15)
*#(*(0(0(x1)))) *#(*(x1)) (16)
*#(*(0(0(x1)))) *#(x1) (17)
*#(*(1(1(x1)))) 0#(0(x1)) (18)
*#(*(1(1(x1)))) 0#(x1) (19)
0#(0(#(#(x1)))) 0#(0(x1)) (20)
0#(0(#(#(x1)))) 0#(x1) (21)
1#(1(#(#(x1)))) 1#(1(x1)) (22)
1#(1(#(#(x1)))) 1#(x1) (23)
$#($(#(#(x1)))) $#($(*(*(x1)))) (24)
$#($(#(#(x1)))) $#(*(*(x1))) (25)
$#($(#(#(x1)))) *#(*(x1)) (26)
$#($(#(#(x1)))) *#(x1) (27)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.