Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/turing_add)

The rewrite relation of the following TRS is considered.

1(q0(1(x1))) 0(1(q1(x1))) (1)
1(q0(0(x1))) 0(0(q1(x1))) (2)
1(q1(1(x1))) 1(1(q1(x1))) (3)
1(q1(0(x1))) 1(0(q1(x1))) (4)
0(q1(x1)) q2(1(x1)) (5)
1(q2(x1)) q2(1(x1)) (6)
0(q2(x1)) 0(q0(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
1(q0(1(x1))) q1(1(0(x1))) (8)
0(q0(1(x1))) q1(0(0(x1))) (9)
1(q1(1(x1))) q1(1(1(x1))) (10)
0(q1(1(x1))) q1(0(1(x1))) (11)
q1(0(x1)) 1(q2(x1)) (12)
q2(1(x1)) 1(q2(x1)) (13)
q2(0(x1)) q0(0(x1)) (14)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(q0(1(x1))) q1#(1(0(x1))) (15)
1#(q0(1(x1))) 1#(0(x1)) (16)
1#(q0(1(x1))) 0#(x1) (17)
0#(q0(1(x1))) q1#(0(0(x1))) (18)
0#(q0(1(x1))) 0#(0(x1)) (19)
0#(q0(1(x1))) 0#(x1) (20)
1#(q1(1(x1))) q1#(1(1(x1))) (21)
1#(q1(1(x1))) 1#(1(x1)) (22)
0#(q1(1(x1))) q1#(0(1(x1))) (23)
0#(q1(1(x1))) 0#(1(x1)) (24)
q1#(0(x1)) 1#(q2(x1)) (25)
q1#(0(x1)) q2#(x1) (26)
q2#(1(x1)) 1#(q2(x1)) (27)
q2#(1(x1)) q2#(x1) (28)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.