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 AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
*(0(x1)) 1(*(x1)) (8)
*(1(x1)) #(0(x1)) (9)
0(#(x1)) #(0(x1)) (10)
1(#(x1)) #(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(x1)) 1#(*(x1)) (14)
*#(0(x1)) *#(x1) (15)
*#(1(x1)) 0#(x1) (16)
0#(#(x1)) 0#(x1) (17)
1#(#(x1)) 1#(x1) (18)
$#(#(x1)) $#(*(x1)) (19)
$#(#(x1)) *#(x1) (20)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.