Certification Problem

Input (TPDB TRS_Standard/Secret_07_TRS/secret3)

The rewrite relation of the following TRS is considered.

app(nil,k) k (1)
app(l,nil) l (2)
app(cons(x,l),k) cons(x,app(l,k)) (3)
sum(cons(x,nil)) cons(x,nil) (4)
sum(cons(x,cons(y,l))) sum(cons(a(x,y,h),l)) (5)
a(h,h,x) s(x) (6)
a(x,s(y),h) a(x,y,s(h)) (7)
a(x,s(y),s(z)) a(x,y,a(x,s(y),z)) (8)
a(s(x),h,z) a(x,z,z) (9)

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#(x,s(y),h) a#(x,y,s(h)) (10)
a#(x,s(y),s(z)) a#(x,y,a(x,s(y),z)) (11)
a#(x,s(y),s(z)) a#(x,s(y),z) (12)
app#(cons(x,l),k) app#(l,k) (13)
sum#(cons(x,cons(y,l))) sum#(cons(a(x,y,h),l)) (14)
a#(s(x),h,z) a#(x,z,z) (15)
sum#(cons(x,cons(y,l))) a#(x,y,h) (16)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.