Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/PEANO_nokinds_Z)

The rewrite relation of the following TRS is considered.

U11(tt,N) activate(N) (1)
U21(tt,M,N) s(plus(activate(N),activate(M))) (2)
and(tt,X) activate(X) (3)
isNat(n__0) tt (4)
isNat(n__plus(V1,V2)) and(isNat(activate(V1)),n__isNat(activate(V2))) (5)
isNat(n__s(V1)) isNat(activate(V1)) (6)
plus(N,0) U11(isNat(N),N) (7)
plus(N,s(M)) U21(and(isNat(M),n__isNat(N)),M,N) (8)
0 n__0 (9)
plus(X1,X2) n__plus(X1,X2) (10)
isNat(X) n__isNat(X) (11)
s(X) n__s(X) (12)
activate(n__0) 0 (13)
activate(n__plus(X1,X2)) plus(X1,X2) (14)
activate(n__isNat(X)) isNat(X) (15)
activate(n__s(X)) s(X) (16)
activate(X) X (17)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
U11#(tt,N) activate#(N) (18)
U21#(tt,M,N) activate#(M) (19)
U21#(tt,M,N) activate#(N) (20)
U21#(tt,M,N) plus#(activate(N),activate(M)) (21)
U21#(tt,M,N) s#(plus(activate(N),activate(M))) (22)
and#(tt,X) activate#(X) (23)
isNat#(n__plus(V1,V2)) activate#(V2) (24)
isNat#(n__plus(V1,V2)) activate#(V1) (25)
isNat#(n__plus(V1,V2)) isNat#(activate(V1)) (26)
isNat#(n__plus(V1,V2)) and#(isNat(activate(V1)),n__isNat(activate(V2))) (27)
isNat#(n__s(V1)) activate#(V1) (28)
isNat#(n__s(V1)) isNat#(activate(V1)) (29)
plus#(N,0) isNat#(N) (30)
plus#(N,0) U11#(isNat(N),N) (31)
plus#(N,s(M)) isNat#(M) (32)
plus#(N,s(M)) and#(isNat(M),n__isNat(N)) (33)
plus#(N,s(M)) U21#(and(isNat(M),n__isNat(N)),M,N) (34)
activate#(n__0) 0# (35)
activate#(n__plus(X1,X2)) plus#(X1,X2) (36)
activate#(n__isNat(X)) isNat#(X) (37)
activate#(n__s(X)) s#(X) (38)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.