Certification Problem

Input (TPDB TRS_Standard/Secret_07_TRS/secret4)

The rewrite relation of the following TRS is considered.

a(h,h,h,x) s(x) (1)
a(l,x,s(y),h) a(l,x,y,s(h)) (2)
a(l,x,s(y),s(z)) a(l,x,y,a(l,x,s(y),z)) (3)
a(l,s(x),h,z) a(l,x,z,z) (4)
a(s(l),h,h,z) a(l,z,h,z) (5)
+(x,h) x (6)
+(h,x) x (7)
+(s(x),s(y)) s(s(+(x,y))) (8)
+(+(x,y),z) +(x,+(y,z)) (9)
s(h) 1 (10)
app(nil,k) k (11)
app(l,nil) l (12)
app(cons(x,l),k) cons(x,app(l,k)) (13)
sum(cons(x,nil)) cons(x,nil) (14)
sum(cons(x,cons(y,l))) sum(cons(a(x,y,h,h),l)) (15)

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.
a#(h,h,h,x) s#(x) (16)
a#(l,x,s(y),h) s#(h) (17)
a#(l,x,s(y),h) a#(l,x,y,s(h)) (18)
a#(l,x,s(y),s(z)) a#(l,x,s(y),z) (19)
a#(l,x,s(y),s(z)) a#(l,x,y,a(l,x,s(y),z)) (20)
a#(l,s(x),h,z) a#(l,x,z,z) (21)
a#(s(l),h,h,z) a#(l,z,h,z) (22)
+#(s(x),s(y)) +#(x,y) (23)
+#(s(x),s(y)) s#(+(x,y)) (24)
+#(s(x),s(y)) s#(s(+(x,y))) (25)
+#(+(x,y),z) +#(y,z) (26)
+#(+(x,y),z) +#(x,+(y,z)) (27)
app#(cons(x,l),k) app#(l,k) (28)
sum#(cons(x,cons(y,l))) a#(x,y,h,h) (29)
sum#(cons(x,cons(y,l))) sum#(cons(a(x,y,h,h),l)) (30)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.