/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 35 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) TransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) TransformationProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) DependencyGraphProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) TransformationProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) DependencyGraphProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) TransformationProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) DependencyGraphProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) TransformationProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) DependencyGraphProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) TransformationProof [EQUIVALENT, 11 ms]
        (35) QDP
        (36) DependencyGraphProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) TransformationProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) DependencyGraphProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) QDPOrderProof [EQUIVALENT, 944 ms]
        (43) QDP
        (44) DependencyGraphProof [EQUIVALENT, 0 ms]
        (45) TRUE
    (46) QDP
        (47) QDPOrderProof [EQUIVALENT, 95 ms]
        (48) QDP
        (49) DependencyGraphProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) UsableRulesProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (54) YES


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

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

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(0)) -> F(s(0))
   G(x) -> H(x, x)
   S(x) -> H(x, 0)
   S(x) -> H(0, x)
   F(g(x)) -> G(g(f(x)))
   F(g(x)) -> G(f(x))
   F(g(x)) -> F(x)
   G(s(x)) -> S(s(g(x)))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   H(f(x), g(x)) -> F(s(x))
   H(f(x), g(x)) -> S(x)
   S(0) -> K(0)
   S(s(0)) -> K(s(0))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(s(s(0)))))
   K(s(s(0))) -> S(s(s(s(0))))
   K(s(s(0))) -> S(s(s(0)))
   H(k(x), g(x)) -> K(s(x))
   H(k(x), g(x)) -> S(x)

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(g(x)) -> G(g(f(x)))
   G(x) -> H(x, x)
   H(f(x), g(x)) -> F(s(x))
   F(g(x)) -> G(f(x))
   G(s(x)) -> S(s(g(x)))
   S(s(0)) -> F(s(0))
   F(g(x)) -> F(x)
   S(s(0)) -> K(s(0))
   K(s(s(0))) -> S(s(s(s(s(0)))))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(s(0))))
   K(s(s(0))) -> S(s(s(0)))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   H(f(x), g(x)) -> S(x)
   H(k(x), g(x)) -> K(s(x))
   H(k(x), g(x)) -> S(x)

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule S(s(0)) -> F(s(0)) at position [0] we obtained the following new rules [LPAR04]:

   (S(s(0)) -> F(h(0, 0)),S(s(0)) -> F(h(0, 0)))
   (S(s(0)) -> F(k(0)),S(s(0)) -> F(k(0)))


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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(g(x)) -> G(g(f(x)))
   G(x) -> H(x, x)
   H(f(x), g(x)) -> F(s(x))
   F(g(x)) -> G(f(x))
   G(s(x)) -> S(s(g(x)))
   F(g(x)) -> F(x)
   S(s(0)) -> K(s(0))
   K(s(s(0))) -> S(s(s(s(s(0)))))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(s(0))))
   K(s(s(0))) -> S(s(s(0)))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   H(f(x), g(x)) -> S(x)
   H(k(x), g(x)) -> K(s(x))
   H(k(x), g(x)) -> S(x)
   S(s(0)) -> F(h(0, 0))
   S(s(0)) -> F(k(0))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(s(s(s(0)))))
   S(s(0)) -> K(s(0))
   K(s(s(0))) -> S(s(s(s(0))))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule S(s(0)) -> K(s(0)) at position [0] we obtained the following new rules [LPAR04]:

   (S(s(0)) -> K(h(0, 0)),S(s(0)) -> K(h(0, 0)))
   (S(s(0)) -> K(k(0)),S(s(0)) -> K(k(0)))


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(s(s(s(0)))))
   K(s(s(0))) -> S(s(s(s(0))))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(0)))
   S(s(0)) -> K(h(0, 0))
   S(s(0)) -> K(k(0))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(s(s(0)))))
   K(s(s(0))) -> S(s(s(s(0))))
   K(s(s(0))) -> S(s(s(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(14) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(s(s(s(s(0))))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(h(s(s(s(0))), 0)),K(s(s(0))) -> S(h(s(s(s(0))), 0)))
   (K(s(s(0))) -> S(h(0, s(s(s(0))))),K(s(s(0))) -> S(h(0, s(s(s(0))))))
   (K(s(s(0))) -> S(s(h(s(s(0)), 0))),K(s(s(0))) -> S(s(h(s(s(0)), 0))))
   (K(s(s(0))) -> S(s(h(0, s(s(0))))),K(s(s(0))) -> S(s(h(0, s(s(0))))))
   (K(s(s(0))) -> S(s(s(f(s(0))))),K(s(s(0))) -> S(s(s(f(s(0))))))
   (K(s(s(0))) -> S(s(s(h(s(0), 0)))),K(s(s(0))) -> S(s(s(h(s(0), 0)))))
   (K(s(s(0))) -> S(s(s(h(0, s(0))))),K(s(s(0))) -> S(s(s(h(0, s(0))))))
   (K(s(s(0))) -> S(s(s(k(s(0))))),K(s(s(0))) -> S(s(s(k(s(0))))))
   (K(s(s(0))) -> S(s(s(s(h(0, 0))))),K(s(s(0))) -> S(s(s(s(h(0, 0))))))
   (K(s(s(0))) -> S(s(s(s(k(0))))),K(s(s(0))) -> S(s(s(s(k(0))))))


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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(s(0))))
   K(s(s(0))) -> S(s(s(0)))
   K(s(s(0))) -> S(h(s(s(s(0))), 0))
   K(s(s(0))) -> S(h(0, s(s(s(0)))))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(16) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(s(s(0))))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(18) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(s(s(s(0)))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(h(s(s(0)), 0)),K(s(s(0))) -> S(h(s(s(0)), 0)))
   (K(s(s(0))) -> S(h(0, s(s(0)))),K(s(s(0))) -> S(h(0, s(s(0)))))
   (K(s(s(0))) -> S(s(f(s(0)))),K(s(s(0))) -> S(s(f(s(0)))))
   (K(s(s(0))) -> S(s(h(s(0), 0))),K(s(s(0))) -> S(s(h(s(0), 0))))
   (K(s(s(0))) -> S(s(h(0, s(0)))),K(s(s(0))) -> S(s(h(0, s(0)))))
   (K(s(s(0))) -> S(s(k(s(0)))),K(s(s(0))) -> S(s(k(s(0)))))
   (K(s(s(0))) -> S(s(s(h(0, 0)))),K(s(s(0))) -> S(s(s(h(0, 0)))))
   (K(s(s(0))) -> S(s(s(k(0)))),K(s(s(0))) -> S(s(s(k(0)))))


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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(h(s(s(0)), 0))
   K(s(s(0))) -> S(h(0, s(s(0))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(20) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(s(0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(22) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(s(s(0))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(f(s(0))),K(s(s(0))) -> S(f(s(0))))
   (K(s(s(0))) -> S(h(s(0), 0)),K(s(s(0))) -> S(h(s(0), 0)))
   (K(s(s(0))) -> S(h(0, s(0))),K(s(s(0))) -> S(h(0, s(0))))
   (K(s(s(0))) -> S(k(s(0))),K(s(s(0))) -> S(k(s(0))))
   (K(s(s(0))) -> S(s(h(0, 0))),K(s(s(0))) -> S(s(h(0, 0))))
   (K(s(s(0))) -> S(s(k(0))),K(s(s(0))) -> S(s(k(0))))


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

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(f(s(0)))
   K(s(s(0))) -> S(h(s(0), 0))
   K(s(s(0))) -> S(h(0, s(0)))
   K(s(s(0))) -> S(k(s(0)))
   K(s(s(0))) -> S(s(h(0, 0)))
   K(s(s(0))) -> S(s(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(24) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(f(s(0)))
   K(s(s(0))) -> S(k(s(0)))
   K(s(s(0))) -> S(s(h(0, 0)))
   K(s(s(0))) -> S(s(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(f(s(0))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(f(h(0, 0))),K(s(s(0))) -> S(f(h(0, 0))))
   (K(s(s(0))) -> S(f(k(0))),K(s(s(0))) -> S(f(k(0))))


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(k(s(0)))
   K(s(s(0))) -> S(s(h(0, 0)))
   K(s(s(0))) -> S(s(k(0)))
   K(s(s(0))) -> S(f(h(0, 0)))
   K(s(s(0))) -> S(f(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(28) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(k(s(0)))
   K(s(s(0))) -> S(s(h(0, 0)))
   K(s(s(0))) -> S(s(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(30) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(k(s(0))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(0),K(s(s(0))) -> S(0))
   (K(s(s(0))) -> S(k(h(0, 0))),K(s(s(0))) -> S(k(h(0, 0))))
   (K(s(s(0))) -> S(k(k(0))),K(s(s(0))) -> S(k(k(0))))


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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(s(h(0, 0)))
   K(s(s(0))) -> S(s(k(0)))
   K(s(s(0))) -> S(0)
   K(s(s(0))) -> S(k(h(0, 0)))
   K(s(s(0))) -> S(k(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(32) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(s(h(0, 0)))
   K(s(s(0))) -> S(s(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(34) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(s(h(0, 0))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(h(h(0, 0), 0)),K(s(s(0))) -> S(h(h(0, 0), 0)))
   (K(s(s(0))) -> S(h(0, h(0, 0))),K(s(s(0))) -> S(h(0, h(0, 0))))


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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(s(k(0)))
   K(s(s(0))) -> S(h(h(0, 0), 0))
   K(s(s(0))) -> S(h(0, h(0, 0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(36) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(s(k(0)))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(38) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule K(s(s(0))) -> S(s(k(0))) at position [0] we obtained the following new rules [LPAR04]:

   (K(s(s(0))) -> S(h(k(0), 0)),K(s(s(0))) -> S(h(k(0), 0)))
   (K(s(s(0))) -> S(h(0, k(0))),K(s(s(0))) -> S(h(0, k(0))))
   (K(s(s(0))) -> S(0),K(s(s(0))) -> S(0))


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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
   K(s(s(0))) -> S(h(k(0), 0))
   K(s(s(0))) -> S(h(0, k(0)))
   K(s(s(0))) -> S(0)

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(40) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(42) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   K(s(s(0))) -> S(s(h(s(s(0)), 0)))
   S(s(s(s(s(s(s(0))))))) -> K(s(s(0)))
   K(s(s(0))) -> S(s(h(0, s(s(0)))))
   K(s(s(0))) -> S(s(s(f(s(0)))))
   K(s(s(0))) -> S(s(s(k(s(0)))))
   K(s(s(0))) -> S(s(s(s(h(0, 0)))))
   K(s(s(0))) -> S(s(s(s(k(0)))))
   K(s(s(0))) -> S(s(f(s(0))))
   K(s(s(0))) -> S(s(h(s(0), 0)))
   K(s(s(0))) -> S(s(h(0, s(0))))
   K(s(s(0))) -> S(s(k(s(0))))
   K(s(s(0))) -> S(s(s(h(0, 0))))
   K(s(s(0))) -> S(s(s(k(0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(K(x_1)) =  	[[2]] 	 +  	[[0, 2]] 	* 	x_1
>>>

   <<<
 POL(s(x_1)) =  	[[0], [1]] 	 +  	[[0, 1], [1, 0]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0], [0]]
>>>

   <<<
 POL(S(x_1)) =  	[[0]] 	 +  	[[2, 0]] 	* 	x_1
>>>

   <<<
 POL(h(x_1, x_2)) =  	[[0], [0]] 	 +  	[[0, 1], [0, 0]] 	* 	x_1 	 +  	[[0, 1], [1, 0]] 	* 	x_2
>>>

   <<<
 POL(f(x_1)) =  	[[0], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(k(x_1)) =  	[[0], [0]] 	 +  	[[2, 0], [3, 0]] 	* 	x_1
>>>

   <<<
 POL(g(x_1)) =  	[[0], [0]] 	 +  	[[3, 3], [2, 2]] 	* 	x_1
>>>


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   s(0) -> k(0)
   f(g(x)) -> g(g(f(x)))
   g(x) -> h(x, x)
   h(f(x), g(x)) -> f(s(x))
   h(k(x), g(x)) -> k(s(x))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   s(s(0)) -> f(s(0))
   s(s(0)) -> k(s(0))
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   g(s(x)) -> s(s(g(x)))
   s(k(0)) -> 0
   k(s(0)) -> 0


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

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   K(s(s(0))) -> S(s(s(h(s(0), 0))))
   K(s(s(0))) -> S(s(s(h(0, s(0)))))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(44) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(45)
TRUE

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

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> H(x, x)
   H(f(x), g(x)) -> F(s(x))
   F(g(x)) -> G(g(f(x)))
   G(s(x)) -> G(x)
   F(g(x)) -> G(f(x))
   F(g(x)) -> F(x)

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(47) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   H(f(x), g(x)) -> F(s(x))
   F(g(x)) -> F(x)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( F_1(x_1) ) = max{0, x_1 - 1}
POL( G_1(x_1) ) = 1
POL( f_1(x_1) ) = 2x_1
POL( g_1(x_1) ) = x_1 + 2
POL( h_2(x_1, x_2) ) = 0
POL( s_1(x_1) ) = max{0, -2}
POL( k_1(x_1) ) = max{0, -2}
POL( 0 ) = 0
POL( H_2(x_1, x_2) ) = 1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f(g(x)) -> g(g(f(x)))
   g(x) -> h(x, x)
   h(f(x), g(x)) -> f(s(x))
   h(k(x), g(x)) -> k(s(x))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   s(s(0)) -> f(s(0))
   s(s(0)) -> k(s(0))
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   g(s(x)) -> s(s(g(x)))
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   s(0) -> k(0)
   s(k(0)) -> 0
   k(s(0)) -> 0


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

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> H(x, x)
   F(g(x)) -> G(g(f(x)))
   G(s(x)) -> G(x)
   F(g(x)) -> G(f(x))

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(49) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(50)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(s(x)) -> G(x)

The TRS R consists of the following rules:

   s(s(0)) -> f(s(0))
   g(x) -> h(x, x)
   s(x) -> h(x, 0)
   s(x) -> h(0, x)
   f(g(x)) -> g(g(f(x)))
   g(s(x)) -> s(s(g(x)))
   h(f(x), g(x)) -> f(s(x))
   s(0) -> k(0)
   s(k(0)) -> 0
   s(s(0)) -> k(s(0))
   k(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> k(s(s(0)))
   k(s(s(0))) -> s(s(s(s(s(0)))))
   h(k(x), g(x)) -> k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(51) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(s(x)) -> G(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(53) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*G(s(x)) -> G(x)
The graph contains the following edges 1 > 1


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

(54)
YES