YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 31 ms]
(2) QTRS
(3) RisEmptyProof [EQUIVALENT, 0 ms]
(4) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   fst(0, Z) -> nil
   fst(s, cons(Y)) -> cons(Y)
   from(X) -> cons(X)
   add(0, X) -> X
   add(s, Y) -> s
   len(nil) -> 0
   len(cons(X)) -> s

The set Q consists of the following terms:

   fst(0, x0)
   fst(s, cons(x0))
   from(x0)
   add(0, x0)
   add(s, x0)
   len(nil)
   len(cons(x0))


----------------------------------------

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:nil > len_1 > s > add_2 > from_1 > cons_1 > 0 > fst_2

and weight map:

   0=2
   nil=1
   s=3
   cons_1=1
   from_1=2
   len_1=1
   fst_2=0
   add_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   fst(0, Z) -> nil
   fst(s, cons(Y)) -> cons(Y)
   from(X) -> cons(X)
   add(0, X) -> X
   add(s, Y) -> s
   len(nil) -> 0
   len(cons(X)) -> s




----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
R is empty.
The set Q consists of the following terms:

   fst(0, x0)
   fst(s, cons(x0))
   from(x0)
   add(0, x0)
   add(s, x0)
   len(nil)
   len(cons(x0))


----------------------------------------

(3) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
----------------------------------------

(4)
YES