Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/PEANO_nosorts_noand_GM)

The rewrite relation of the following TRS is considered.

a__U11(tt,M,N) a__U12(tt,M,N) (1)
a__U12(tt,M,N) s(a__plus(mark(N),mark(M))) (2)
a__plus(N,0) mark(N) (3)
a__plus(N,s(M)) a__U11(tt,M,N) (4)
mark(U11(X1,X2,X3)) a__U11(mark(X1),X2,X3) (5)
mark(U12(X1,X2,X3)) a__U12(mark(X1),X2,X3) (6)
mark(plus(X1,X2)) a__plus(mark(X1),mark(X2)) (7)
mark(tt) tt (8)
mark(s(X)) s(mark(X)) (9)
mark(0) 0 (10)
a__U11(X1,X2,X3) U11(X1,X2,X3) (11)
a__U12(X1,X2,X3) U12(X1,X2,X3) (12)
a__plus(X1,X2) plus(X1,X2) (13)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__U12#(tt,M,N) mark#(N) (14)
mark#(s(X)) mark#(X) (15)
a__U11#(tt,M,N) a__U12#(tt,M,N) (16)
mark#(plus(X1,X2)) mark#(X2) (17)
mark#(U11(X1,X2,X3)) mark#(X1) (18)
a__plus#(N,0) mark#(N) (19)
a__plus#(N,s(M)) a__U11#(tt,M,N) (20)
mark#(plus(X1,X2)) a__plus#(mark(X1),mark(X2)) (21)
mark#(U11(X1,X2,X3)) a__U11#(mark(X1),X2,X3) (22)
mark#(U12(X1,X2,X3)) mark#(X1) (23)
a__U12#(tt,M,N) mark#(M) (24)
a__U12#(tt,M,N) a__plus#(mark(N),mark(M)) (25)
mark#(plus(X1,X2)) mark#(X1) (26)
mark#(U12(X1,X2,X3)) a__U12#(mark(X1),X2,X3) (27)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.