/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 120 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 82 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) DependencyGraphProof [EQUIVALENT, 0 ms]
(16) QDP
(17) TransformationProof [EQUIVALENT, 0 ms]
(18) QDP
(19) DependencyGraphProof [EQUIVALENT, 0 ms]
(20) QDP
(21) TransformationProof [EQUIVALENT, 0 ms]
(22) QDP
(23) DependencyGraphProof [EQUIVALENT, 0 ms]
(24) QDP
(25) TransformationProof [EQUIVALENT, 0 ms]
(26) QDP
(27) DependencyGraphProof [EQUIVALENT, 0 ms]
(28) QDP
(29) TransformationProof [EQUIVALENT, 0 ms]
(30) QDP
(31) TransformationProof [EQUIVALENT, 0 ms]
(32) QDP
(33) DependencyGraphProof [EQUIVALENT, 0 ms]
(34) QDP
(35) TransformationProof [EQUIVALENT, 9 ms]
(36) QDP
(37) DependencyGraphProof [EQUIVALENT, 0 ms]
(38) QDP
(39) TransformationProof [EQUIVALENT, 0 ms]
(40) QDP
(41) DependencyGraphProof [EQUIVALENT, 0 ms]
(42) QDP
(43) QDPOrderProof [EQUIVALENT, 2034 ms]
(44) QDP
(45) QDPOrderProof [EQUIVALENT, 1405 ms]
(46) QDP
(47) DependencyGraphProof [EQUIVALENT, 0 ms]
(48) AND
    (49) QDP
        (50) UsableRulesProof [EQUIVALENT, 0 ms]
        (51) QDP
        (52) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (53) YES
    (54) QDP
        (55) QDPOrderProof [EQUIVALENT, 1310 ms]
        (56) QDP
        (57) QDPOrderProof [EQUIVALENT, 762 ms]
        (58) QDP
        (59) QDPOrderProof [EQUIVALENT, 790 ms]
        (60) QDP
        (61) QDPOrderProof [EQUIVALENT, 592 ms]
        (62) QDP
        (63) DependencyGraphProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) QDPOrderProof [EQUIVALENT, 1527 ms]
        (66) QDP
        (67) DependencyGraphProof [EQUIVALENT, 0 ms]
        (68) QDP
        (69) UsableRulesProof [EQUIVALENT, 0 ms]
        (70) QDP
        (71) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (72) YES


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

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

   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) -> f(0)
   s(s(0)) -> f(s(0))
   f(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> f(s(s(0)))
   f(s(s(0))) -> s(s(s(s(s(0)))))
   f(s(s(x))) -> h(f(x), g(h(x, 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(f(0)),S(s(0)) -> F(f(0)))


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

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

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

The TRS R consists of the following rules:

   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) -> f(0)
   s(s(0)) -> f(s(0))
   f(s(0)) -> 0
   s(s(s(s(s(s(s(0))))))) -> f(s(s(0)))
   f(s(s(0))) -> s(s(s(s(s(0)))))
   f(s(s(x))) -> h(f(x), g(h(x, 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 1 SCC with 2 less nodes.
----------------------------------------

(8)
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(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(g(x)) -> F(x)
   F(s(s(0))) -> S(s(s(s(s(0)))))
   F(s(s(0))) -> S(s(s(s(0))))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   F(s(s(x))) -> H(x, x)

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   F(g(x)) -> G(g(f(x)))
   F(g(x)) -> G(f(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) ) = 2x_1
POL( G_1(x_1) ) = x_1
POL( H_2(x_1, x_2) ) = x_1
POL( S_1(x_1) ) = max{0, x_1 - 2}
POL( h_2(x_1, x_2) ) = x_1
POL( f_1(x_1) ) = 2x_1
POL( g_1(x_1) ) = x_1 + 2
POL( s_1(x_1) ) = x_1
POL( 0 ) = 0

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

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


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

(10)
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))
   G(s(x)) -> S(s(g(x)))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(s(s(s(0)))))
   F(s(s(0))) -> S(s(s(s(0))))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   F(s(s(x))) -> H(x, x)

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

POL( F_1(x_1) ) = 2x_1 + 1
POL( G_1(x_1) ) = 2x_1 + 1
POL( H_2(x_1, x_2) ) = max{0, 2x_2 - 1}
POL( S_1(x_1) ) = 1
POL( s_1(x_1) ) = x_1
POL( h_2(x_1, x_2) ) = x_1
POL( 0 ) = 0
POL( f_1(x_1) ) = 2x_1
POL( g_1(x_1) ) = x_1 + 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) -> f(0)
   h(f(x), g(x)) -> f(s(x))
   f(g(x)) -> g(g(f(x)))
   g(x) -> h(x, x)
   g(s(x)) -> s(s(g(x)))
   s(s(0)) -> f(s(0))
   f(s(s(0))) -> s(s(s(s(s(0)))))
   s(s(s(s(s(s(s(0))))))) -> f(s(s(0)))
   f(s(s(x))) -> h(f(x), g(h(x, x)))
   f(s(0)) -> 0


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

   (G(s(y0)) -> S(h(g(y0), 0)),G(s(y0)) -> S(h(g(y0), 0)))
   (G(s(y0)) -> S(h(0, g(y0))),G(s(y0)) -> S(h(0, g(y0))))
   (G(s(x0)) -> S(s(h(x0, x0))),G(s(x0)) -> S(s(h(x0, x0))))
   (G(s(s(x0))) -> S(s(s(s(g(x0))))),G(s(s(x0))) -> S(s(s(s(g(x0))))))


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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(f(x), g(x)) -> F(s(x))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(s(s(s(0)))))
   F(s(s(0))) -> S(s(s(s(0))))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(y0)) -> S(h(g(y0), 0))
   G(s(y0)) -> S(h(0, g(y0)))
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))

The TRS R consists of the following rules:

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

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

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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(0))) -> S(s(s(s(s(0)))))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(s(s(0))))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   H(f(x), g(x)) -> S(x)

The TRS R consists of the following rules:

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

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

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

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


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(s(s(0))))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   H(f(x), g(x)) -> S(x)
   F(s(s(0))) -> S(h(s(s(s(0))), 0))
   F(s(s(0))) -> S(h(0, s(s(s(0)))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))

The TRS R consists of the following rules:

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

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

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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(0))) -> S(s(s(s(0))))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   H(f(x), g(x)) -> S(x)

The TRS R consists of the following rules:

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

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

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

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


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   H(f(x), g(x)) -> S(x)
   F(s(s(0))) -> S(h(s(s(0)), 0))
   F(s(s(0))) -> S(h(0, s(s(0))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))

The TRS R consists of the following rules:

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

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

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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(0))) -> S(s(s(0)))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   H(f(x), g(x)) -> S(x)

The TRS R consists of the following rules:

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

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

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

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


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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   H(f(x), g(x)) -> S(x)
   F(s(s(0))) -> S(h(s(0), 0))
   F(s(s(0))) -> S(h(0, s(0)))
   F(s(s(0))) -> S(f(s(0)))
   F(s(s(0))) -> S(s(h(0, 0)))
   F(s(s(0))) -> S(s(f(0)))

The TRS R consists of the following rules:

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

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

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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> S(g(x))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   F(s(s(0))) -> S(f(s(0)))
   F(s(s(0))) -> S(s(h(0, 0)))
   F(s(s(0))) -> S(s(f(0)))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   H(f(x), g(x)) -> S(x)

The TRS R consists of the following rules:

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

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

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

   (G(s(x0)) -> S(h(x0, x0)),G(s(x0)) -> S(h(x0, x0)))
   (G(s(s(x0))) -> S(s(s(g(x0)))),G(s(s(x0))) -> S(s(s(g(x0)))))


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

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   F(s(s(0))) -> S(f(s(0)))
   F(s(s(0))) -> S(s(h(0, 0)))
   F(s(s(0))) -> S(s(f(0)))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   H(f(x), g(x)) -> S(x)
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))

The TRS R consists of the following rules:

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

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

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

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


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   F(s(s(0))) -> S(s(h(0, 0)))
   F(s(s(0))) -> S(s(f(0)))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   H(f(x), g(x)) -> S(x)
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))
   F(s(s(0))) -> S(0)
   F(s(s(0))) -> S(f(h(0, 0)))
   F(s(s(0))) -> S(f(f(0)))

The TRS R consists of the following rules:

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

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

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

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   F(s(s(0))) -> S(s(h(0, 0)))
   F(s(s(0))) -> S(s(f(0)))

The TRS R consists of the following rules:

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

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

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

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


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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   F(s(s(0))) -> S(s(f(0)))
   F(s(s(0))) -> S(h(h(0, 0), 0))
   F(s(s(0))) -> S(h(0, h(0, 0)))

The TRS R consists of the following rules:

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

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

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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   F(s(s(0))) -> S(s(f(0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))
   H(f(x), g(x)) -> S(x)

The TRS R consists of the following rules:

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

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

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

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


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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))
   H(f(x), g(x)) -> S(x)
   F(s(s(0))) -> S(h(f(0), 0))
   F(s(s(0))) -> S(h(0, f(0)))

The TRS R consists of the following rules:

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

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

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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   G(s(x0)) -> S(h(x0, x0))
   G(s(s(x0))) -> S(s(s(g(x0))))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   G(s(s(x0))) -> S(s(s(s(g(x0)))))
   G(s(s(x0))) -> S(s(s(g(x0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

   <<<
 POL(g(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[0A, 0A, -I], [-I, 0A, -I], [-I, 0A, 0A]] 	* 	x_1
>>>

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

   <<<
 POL(s(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[0A, 0A, 0A], [-I, 0A, -I], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [-I], [-I]] 	 +  	[[0A, 0A, -I], [-I, 0A, -I], [-I, 0A, -I]] 	* 	x_1 	 +  	[[0A, 0A, -I], [-I, -I, -I], [-I, 0A, -I]] 	* 	x_2
>>>

   <<<
 POL(S(x_1)) =  	[[0A]] 	 +  	[[-I, -I, -I]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A], [-I], [0A]]
>>>

   <<<
 POL(G(x_1)) =  	[[0A]] 	 +  	[[0A, -I, 1A]] 	* 	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) -> f(0)
   h(f(x), g(x)) -> f(s(x))
   f(g(x)) -> g(g(f(x)))
   g(x) -> h(x, x)
   g(s(x)) -> s(s(g(x)))
   s(s(0)) -> f(s(0))
   f(s(s(0))) -> s(s(s(s(s(0)))))
   s(s(s(s(s(s(s(0))))))) -> f(s(s(0)))
   f(s(s(x))) -> h(f(x), g(h(x, x)))
   f(s(0)) -> 0


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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(f(x), g(x)) -> F(s(x))
   F(s(s(x))) -> H(f(x), g(h(x, x)))
   H(f(x), g(x)) -> S(x)
   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
   F(s(s(x))) -> F(x)
   F(s(s(x))) -> G(h(x, x))
   G(s(x)) -> G(x)
   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(x0)) -> S(h(x0, x0))
   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(f(s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(s(s(f(0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
   F(s(s(0))) -> S(s(s(f(0))))

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   G(s(x0)) -> S(s(h(x0, x0)))
   G(s(x0)) -> S(h(x0, x0))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(H(x_1, x_2)) =  	[[0A]] 	 +  	[[-I, -I, -I]] 	* 	x_1 	 +  	[[-I, 0A, -I]] 	* 	x_2
>>>

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

   <<<
 POL(g(x_1)) =  	[[1A], [0A], [1A]] 	 +  	[[0A, 0A, -I], [0A, 0A, -I], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[0A, -I, -I]] 	* 	x_1
>>>

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

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [0A], [0A]] 	 +  	[[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] 	* 	x_1 	 +  	[[-I, -I, -I], [0A, -I, -I], [-I, -I, -I]] 	* 	x_2
>>>

   <<<
 POL(S(x_1)) =  	[[0A]] 	 +  	[[0A, -I, -I]] 	* 	x_1
>>>

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

   <<<
 POL(G(x_1)) =  	[[0A]] 	 +  	[[0A, 0A, -I]] 	* 	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) -> f(0)
   h(f(x), g(x)) -> f(s(x))
   f(g(x)) -> g(g(f(x)))
   g(x) -> h(x, x)
   g(s(x)) -> s(s(g(x)))
   s(s(0)) -> f(s(0))
   f(s(s(0))) -> s(s(s(s(s(0)))))
   s(s(s(s(s(s(s(0))))))) -> f(s(s(0)))
   f(s(s(x))) -> h(f(x), g(h(x, x)))
   f(s(0)) -> 0


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

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

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

The TRS R consists of the following rules:

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

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

(47) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(48)
Complex Obligation (AND)

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

(49)
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:

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

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

(50) 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.
----------------------------------------

(51)
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.
----------------------------------------

(52) 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


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

(53)
YES

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

(54)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   F(s(s(0))) -> S(s(h(s(s(0)), 0)))
   F(s(s(0))) -> S(s(h(0, s(s(0)))))
   F(s(s(0))) -> S(s(s(h(s(0), 0))))
   F(s(s(0))) -> S(s(s(h(0, s(0)))))
   F(s(s(0))) -> S(s(s(s(h(0, 0)))))
   F(s(s(0))) -> S(s(h(s(0), 0)))
   F(s(s(0))) -> S(s(h(0, s(0))))
   F(s(s(0))) -> S(s(s(h(0, 0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[0A, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(s(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[0A, 0A, -I], [0A, 0A, -I], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(H(x_1, x_2)) =  	[[0A]] 	 +  	[[0A, 0A, 0A]] 	* 	x_1 	 +  	[[0A, 0A, 0A]] 	* 	x_2
>>>

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

   <<<
 POL(g(x_1)) =  	[[0A], [0A], [-I]] 	 +  	[[0A, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [0A], [-I]] 	 +  	[[0A, 0A, -I], [0A, 0A, -I], [0A, -I, -I]] 	* 	x_1 	 +  	[[0A, 0A, -I], [0A, 0A, -I], [0A, -I, -I]] 	* 	x_2
>>>

   <<<
 POL(0) =  	[[0A], [0A], [1A]]
>>>

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


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

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


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

(56)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   F(s(s(0))) -> S(s(f(s(0))))
   F(s(s(0))) -> S(s(s(f(0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(s(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[-I, -I, -I], [0A, -I, 0A], [0A, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(H(x_1, x_2)) =  	[[0A]] 	 +  	[[0A, 0A, 0A]] 	* 	x_1 	 +  	[[-I, -I, -I]] 	* 	x_2
>>>

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

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

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [0A], [0A]] 	 +  	[[-I, -I, -I], [0A, -I, -I], [-I, -I, -I]] 	* 	x_1 	 +  	[[-I, -I, -I], [0A, -I, -I], [0A, -I, -I]] 	* 	x_2
>>>

   <<<
 POL(S(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A], [1A], [0A]]
>>>


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

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


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

(58)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   F(s(s(0))) -> S(s(s(s(f(0)))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, 1A]] 	* 	x_1
>>>

   <<<
 POL(s(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[0A, -I, 0A], [-I, 0A, 0A], [-I, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(H(x_1, x_2)) =  	[[0A]] 	 +  	[[0A, -I, -I]] 	* 	x_1 	 +  	[[0A, 1A, 0A]] 	* 	x_2
>>>

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

   <<<
 POL(g(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, 0A, 0A], [-I, 0A, 0A], [-I, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, -I, 0A], [-I, -I, -I], [-I, 0A, -I]] 	* 	x_1 	 +  	[[-I, -I, 0A], [-I, 0A, -I], [-I, 0A, -I]] 	* 	x_2
>>>

   <<<
 POL(S(x_1)) =  	[[0A]] 	 +  	[[-I, -I, 1A]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A], [-I], [0A]]
>>>


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

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


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

(60)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   S(s(s(s(s(s(s(0))))))) -> F(s(s(0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, -I]] 	* 	x_1
>>>

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

   <<<
 POL(H(x_1, x_2)) =  	[[0A]] 	 +  	[[-I, 0A, -I]] 	* 	x_1 	 +  	[[0A, 0A, 0A]] 	* 	x_2
>>>

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

   <<<
 POL(g(x_1)) =  	[[0A], [0A], [-I]] 	 +  	[[0A, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [0A], [-I]] 	 +  	[[-I, 0A, 0A], [-I, -I, 0A], [0A, 0A, 0A]] 	* 	x_1 	 +  	[[-I, 0A, 0A], [-I, -I, 0A], [0A, 0A, 0A]] 	* 	x_2
>>>

   <<<
 POL(S(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A], [-I], [-I]]
>>>


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

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


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

(62)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

(64)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   F(s(s(x))) -> H(f(x), g(h(x, x)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(H(x_1, x_2)) =  	[[0A]] 	 +  	[[-I, 0A, -I]] 	* 	x_1 	 +  	[[0A, -I, -I]] 	* 	x_2
>>>

   <<<
 POL(f(x_1)) =  	[[1A], [-I], [1A]] 	 +  	[[-I, 1A, -I], [-I, 0A, -I], [-I, 1A, -I]] 	* 	x_1
>>>

   <<<
 POL(g(x_1)) =  	[[0A], [1A], [2A]] 	 +  	[[0A, 1A, 0A], [-I, 0A, -I], [-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(F(x_1)) =  	[[0A]] 	 +  	[[-I, 1A, 0A]] 	* 	x_1
>>>

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

   <<<
 POL(h(x_1, x_2)) =  	[[0A], [-I], [0A]] 	 +  	[[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] 	* 	x_1 	 +  	[[-I, 0A, -I], [-I, -I, -I], [-I, 0A, -I]] 	* 	x_2
>>>

   <<<
 POL(0) =  	[[0A], [-I], [1A]]
>>>


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


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

(66)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(67) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(68)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> F(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(69) 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.
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(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> F(x)

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

(71) 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:
*F(s(s(x))) -> F(x)
The graph contains the following edges 1 > 1


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(72)
YES