Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/MYNAT_nosorts_GM)

The rewrite relation of the following TRS is considered.

a__and(tt,X) mark(X) (1)
a__plus(N,0) mark(N) (2)
a__plus(N,s(M)) s(a__plus(mark(N),mark(M))) (3)
a__x(N,0) 0 (4)
a__x(N,s(M)) a__plus(a__x(mark(N),mark(M)),mark(N)) (5)
mark(and(X1,X2)) a__and(mark(X1),X2) (6)
mark(plus(X1,X2)) a__plus(mark(X1),mark(X2)) (7)
mark(x(X1,X2)) a__x(mark(X1),mark(X2)) (8)
mark(tt) tt (9)
mark(0) 0 (10)
mark(s(X)) s(mark(X)) (11)
a__and(X1,X2) and(X1,X2) (12)
a__plus(X1,X2) plus(X1,X2) (13)
a__x(X1,X2) x(X1,X2) (14)

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.
mark#(and(X1,X2)) a__and#(mark(X1),X2) (15)
mark#(plus(X1,X2)) mark#(X1) (16)
a__plus#(N,s(M)) a__plus#(mark(N),mark(M)) (17)
a__x#(N,s(M)) a__x#(mark(N),mark(M)) (18)
a__x#(N,s(M)) mark#(M) (19)
a__x#(N,s(M)) mark#(N) (20)
a__plus#(N,s(M)) mark#(N) (21)
mark#(plus(X1,X2)) mark#(X2) (22)
a__x#(N,s(M)) mark#(N) (20)
mark#(plus(X1,X2)) a__plus#(mark(X1),mark(X2)) (23)
mark#(x(X1,X2)) a__x#(mark(X1),mark(X2)) (24)
mark#(x(X1,X2)) mark#(X2) (25)
a__and#(tt,X) mark#(X) (26)
mark#(and(X1,X2)) mark#(X1) (27)
a__plus#(N,0) mark#(N) (28)
a__x#(N,s(M)) a__plus#(a__x(mark(N),mark(M)),mark(N)) (29)
a__plus#(N,s(M)) mark#(M) (30)
mark#(s(X)) mark#(X) (31)
mark#(x(X1,X2)) mark#(X1) (32)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.