Certification Problem

Input (TPDB TRS_Standard/Secret_07_TRS/secret5)

The rewrite relation of the following TRS is considered.

t(N) cs(r(q(N)),nt(ns(N))) (1)
q(0) 0 (2)
q(s(X)) s(p(q(X),d(X))) (3)
d(0) 0 (4)
d(s(X)) s(s(d(X))) (5)
p(0,X) X (6)
p(X,0) X (7)
p(s(X),s(Y)) s(s(p(X,Y))) (8)
f(0,X) nil (9)
f(s(X),cs(Y,Z)) cs(Y,nf(X,a(Z))) (10)
t(X) nt(X) (11)
s(X) ns(X) (12)
f(X1,X2) nf(X1,X2) (13)
a(nt(X)) t(a(X)) (14)
a(ns(X)) s(a(X)) (15)
a(nf(X1,X2)) f(a(X1),a(X2)) (16)
a(X) X (17)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
t#(N) q#(N) (18)
q#(s(X)) s#(p(q(X),d(X))) (19)
q#(s(X)) p#(q(X),d(X)) (20)
q#(s(X)) q#(X) (21)
q#(s(X)) d#(X) (22)
d#(s(X)) s#(s(d(X))) (23)
d#(s(X)) s#(d(X)) (24)
d#(s(X)) d#(X) (25)
p#(s(X),s(Y)) s#(s(p(X,Y))) (26)
p#(s(X),s(Y)) s#(p(X,Y)) (27)
p#(s(X),s(Y)) p#(X,Y) (28)
f#(s(X),cs(Y,Z)) a#(Z) (29)
a#(nt(X)) t#(a(X)) (30)
a#(nt(X)) a#(X) (31)
a#(ns(X)) s#(a(X)) (32)
a#(ns(X)) a#(X) (33)
a#(nf(X1,X2)) f#(a(X1),a(X2)) (34)
a#(nf(X1,X2)) a#(X1) (35)
a#(nf(X1,X2)) a#(X2) (36)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.