Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex8_BLR02_FR)

The rewrite relation of the following TRS is considered.

fib(N) sel(N,fib1(s(0),s(0))) (1)
fib1(X,Y) cons(X,n__fib1(Y,n__add(X,Y))) (2)
add(0,X) X (3)
add(s(X),Y) s(add(X,Y)) (4)
sel(0,cons(X,XS)) X (5)
sel(s(N),cons(X,XS)) sel(N,activate(XS)) (6)
fib1(X1,X2) n__fib1(X1,X2) (7)
add(X1,X2) n__add(X1,X2) (8)
activate(n__fib1(X1,X2)) fib1(activate(X1),activate(X2)) (9)
activate(n__add(X1,X2)) add(activate(X1),activate(X2)) (10)
activate(X) X (11)

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.
sel#(s(N),cons(X,XS)) activate#(XS) (12)
activate#(n__add(X1,X2)) add#(activate(X1),activate(X2)) (13)
fib#(N) sel#(N,fib1(s(0),s(0))) (14)
add#(s(X),Y) add#(X,Y) (15)
activate#(n__add(X1,X2)) activate#(X1) (16)
fib#(N) fib1#(s(0),s(0)) (17)
activate#(n__fib1(X1,X2)) activate#(X2) (18)
activate#(n__fib1(X1,X2)) fib1#(activate(X1),activate(X2)) (19)
sel#(s(N),cons(X,XS)) sel#(N,activate(XS)) (20)
activate#(n__add(X1,X2)) activate#(X2) (21)
activate#(n__fib1(X1,X2)) activate#(X1) (22)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.