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

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

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

i(x, x) → i(a, b)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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:

I(x, x) → I(a, b)
G(x, x) → G(a, b)
H(s(f(x))) → H(f(x))
F(s(x)) → F(h(s(x)))
F(s(x)) → H(s(x))
F(g(s(x), y)) → F(g(x, s(y)))
F(g(s(x), y)) → G(x, s(y))
H(g(x, s(y))) → H(g(s(x), y))
H(g(x, s(y))) → G(s(x), y)
H(i(x, y)) → I(i(c, h(h(y))), x)
H(i(x, y)) → I(c, h(h(y)))
H(i(x, y)) → H(h(y))
H(i(x, y)) → H(y)
G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(a, g(x, g(b, g(b, y))))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(x, g(b, g(b, y)))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(x, g(b, g(b, y))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(x, g(b, g(a, g(x, y))))) → G(b, g(b, y))
G(a, g(x, g(b, g(a, g(x, y))))) → G(b, y)

The TRS R consists of the following rules:

i(x, x) → i(a, b)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(x, g(b, g(b, y))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(x, g(b, g(b, y)))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))

The TRS R consists of the following rules:

i(x, x) → i(a, b)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(6) UsableRulesProof (EQUIVALENT transformation)

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.

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(x, g(b, g(b, y))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(x, g(b, g(b, y)))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(8) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(x, g(b, g(b, y)))) at position [1] we obtained the following new rules [LPAR04]:

G(a, g(g(b, g(b, y1)), g(b, g(a, g(g(b, g(b, y1)), y1))))) → G(a, g(a, b))
G(a, g(a, g(b, g(a, g(a, g(a, g(b, x1))))))) → G(a, g(a, g(a, g(a, g(b, g(b, g(b, x1)))))))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(y0, g(b, g(a, b))))

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(x, g(b, g(b, y)))))
G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(g(b, g(b, y1)), g(b, g(a, g(g(b, g(b, y1)), y1))))) → G(a, g(a, b))
G(a, g(a, g(b, g(a, g(a, g(a, g(b, x1))))))) → G(a, g(a, g(a, g(a, g(b, g(b, g(b, x1)))))))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(y0, g(b, g(a, b))))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(10) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(x, g(b, g(b, y)))))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(y0, g(b, g(a, b))))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(12) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(a, g(x, g(b, g(a, g(x, y))))) → G(a, g(a, g(x, g(b, g(b, y))))) at position [1] we obtained the following new rules [LPAR04]:

G(a, g(g(b, g(b, y1)), g(b, g(a, g(g(b, g(b, y1)), y1))))) → G(a, g(a, g(a, b)))
G(a, g(a, g(b, g(a, g(a, g(a, g(b, x1))))))) → G(a, g(a, g(a, g(a, g(a, g(b, g(b, g(b, x1))))))))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(a, g(y0, g(b, g(a, b)))))

(13) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(y0, g(b, g(a, b))))
G(a, g(g(b, g(b, y1)), g(b, g(a, g(g(b, g(b, y1)), y1))))) → G(a, g(a, g(a, b)))
G(a, g(a, g(b, g(a, g(a, g(a, g(b, x1))))))) → G(a, g(a, g(a, g(a, g(a, g(b, g(b, g(b, x1))))))))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(a, g(y0, g(b, g(a, b)))))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(y0, g(b, g(a, b))))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(a, g(y0, g(b, g(a, b)))))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(16) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(y0, g(b, g(a, b)))) at position [1] we obtained the following new rules [LPAR04]:

G(a, g(g(b, g(a, b)), g(b, g(a, g(g(b, g(a, b)), b))))) → G(a, g(a, b))

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(a, g(y0, g(b, g(a, b)))))
G(a, g(g(b, g(a, b)), g(b, g(a, g(g(b, g(a, b)), b))))) → G(a, g(a, b))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(19) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(a, g(y0, g(b, g(a, b)))))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(20) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule G(a, g(y0, g(b, g(a, g(y0, b))))) → G(a, g(a, g(y0, g(b, g(a, b))))) at position [1] we obtained the following new rules [LPAR04]:

G(a, g(g(b, g(a, b)), g(b, g(a, g(g(b, g(a, b)), b))))) → G(a, g(a, g(a, b)))

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))
G(a, g(g(b, g(a, b)), g(b, g(a, g(g(b, g(a, b)), b))))) → G(a, g(a, g(a, b)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(22) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(24) UsableRulesProof (EQUIVALENT transformation)

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.

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)

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

(26) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule G(a, g(x, g(b, g(a, g(x, y))))) → G(x, g(b, g(b, y))) we obtained the following new rules [LPAR04]:

G(a, g(a, g(b, g(a, g(a, x1))))) → G(a, g(b, g(b, x1)))

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(a, g(a, g(b, g(a, g(a, x1))))) → G(a, g(b, g(b, x1)))

The TRS R consists of the following rules:

g(x, x) → g(a, b)

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

(28) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(29) TRUE

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(g(x, s(y))) → H(g(s(x), y))
H(s(f(x))) → H(f(x))
H(i(x, y)) → H(h(y))
H(i(x, y)) → H(y)

The TRS R consists of the following rules:

i(x, x) → i(a, b)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(31) UsableRulesProof (EQUIVALENT transformation)

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.

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(g(x, s(y))) → H(g(s(x), y))
H(s(f(x))) → H(f(x))
H(i(x, y)) → H(h(y))
H(i(x, y)) → H(y)

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


H(s(f(x))) → H(f(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(H(x1)) =
/0\
\1/
 +
/01\
\00/
·x1

POL(g(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/0\
\1/
 +
/00\
\10/
·x1

POL(f(x1)) =
/1\
\0/
 +
/01\
\01/
·x1

POL(i(x1, x2)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/00\
\01/
·x2

POL(h(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

POL(c) =
/0\
\0/

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

h(i(x, y)) → i(i(c, h(h(y))), x)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
i(x, x) → i(a, b)
f(g(s(x), y)) → f(g(x, s(y)))
f(s(x)) → s(s(f(h(s(x)))))

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(g(x, s(y))) → H(g(s(x), y))
H(i(x, y)) → H(h(y))
H(i(x, y)) → H(y)

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


H(g(x, s(y))) → H(g(s(x), y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(H(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(g(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\11/
·x2

POL(s(x1)) =
/1\
\0/
 +
/01\
\10/
·x1

POL(i(x1, x2)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/00\
\01/
·x2

POL(h(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

POL(c) =
/0\
\0/

POL(f(x1)) =
/0\
\0/
 +
/10\
\10/
·x1

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

h(i(x, y)) → i(i(c, h(h(y))), x)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
i(x, x) → i(a, b)
f(g(s(x), y)) → f(g(x, s(y)))
f(s(x)) → s(s(f(h(s(x)))))

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(i(x, y)) → H(h(y))
H(i(x, y)) → H(y)

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(37) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule H(i(x, y)) → H(h(y)) at position [0] we obtained the following new rules [LPAR04]:

H(i(y0, s(f(x0)))) → H(h(f(x0)))
H(i(y0, g(x0, s(x1)))) → H(h(g(s(x0), x1)))
H(i(y0, i(x0, x1))) → H(i(i(c, h(h(x1))), x0))

(38) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(i(x, y)) → H(y)
H(i(y0, s(f(x0)))) → H(h(f(x0)))
H(i(y0, g(x0, s(x1)))) → H(h(g(s(x0), x1)))
H(i(y0, i(x0, x1))) → H(i(i(c, h(h(x1))), x0))

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


H(i(y0, s(f(x0)))) → H(h(f(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(H(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

POL(i(x1, x2)) =
/0\
\0/
 +
/11\
\01/
·x1 +
/00\
\01/
·x2

POL(s(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

POL(f(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(h(x1)) =
/0\
\0/
 +
/00\
\10/
·x1

POL(g(x1, x2)) =
/0\
\0/
 +
/10\
\00/
·x1 +
/00\
\10/
·x2

POL(c) =
/0\
\0/

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

h(i(x, y)) → i(i(c, h(h(y))), x)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
i(x, x) → i(a, b)
f(g(s(x), y)) → f(g(x, s(y)))
f(s(x)) → s(s(f(h(s(x)))))

(40) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(i(x, y)) → H(y)
H(i(y0, g(x0, s(x1)))) → H(h(g(s(x0), x1)))
H(i(y0, i(x0, x1))) → H(i(i(c, h(h(x1))), x0))

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


H(i(y0, g(x0, s(x1)))) → H(h(g(s(x0), x1)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(H(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(i(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/00\
\11/
·x2

POL(g(x1, x2)) =
/0\
\1/
 +
/01\
\00/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(h(x1)) =
/0\
\0/
 +
/00\
\10/
·x1

POL(c) =
/0\
\0/

POL(f(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

h(i(x, y)) → i(i(c, h(h(y))), x)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
i(x, x) → i(a, b)
f(g(s(x), y)) → f(g(x, s(y)))
f(s(x)) → s(s(f(h(s(x)))))

(42) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(i(x, y)) → H(y)
H(i(y0, i(x0, x1))) → H(i(i(c, h(h(x1))), x0))

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


H(i(x, y)) → H(y)
H(i(y0, i(x0, x1))) → H(i(i(c, h(h(x1))), x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(H(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(i(x1, x2)) =
/1\
\1/
 +
/11\
\00/
·x1 +
/00\
\11/
·x2

POL(c) =
/0\
\0/

POL(h(x1)) =
/0\
\0/
 +
/03\
\10/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(f(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(g(x1, x2)) =
/0\
\0/
 +
/31\
\01/
·x1 +
/00\
\00/
·x2

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

h(i(x, y)) → i(i(c, h(h(y))), x)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
i(x, x) → i(a, b)
f(g(s(x), y)) → f(g(x, s(y)))
f(s(x)) → s(s(f(h(s(x)))))

(44) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
i(x, x) → i(a, b)
g(x, x) → g(a, b)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))

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

(45) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(46) TRUE

(47) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(g(s(x), y)) → F(g(x, s(y)))
F(s(x)) → F(h(s(x)))

The TRS R consists of the following rules:

i(x, x) → i(a, b)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
h(g(x, s(y))) → h(g(s(x), y))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(a, g(x, g(b, g(a, g(x, y))))) → g(a, g(a, g(a, g(x, g(b, g(b, y))))))

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

(48) UsableRulesProof (EQUIVALENT transformation)

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.

(49) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(g(s(x), y)) → F(g(x, s(y)))
F(s(x)) → F(h(s(x)))

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
h(i(x, y)) → i(i(c, h(h(y))), x)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
i(x, x) → i(a, b)

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(g(s(x), y)) → F(g(x, s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(F(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(g(x1, x2)) =
/0\
\0/
 +
/00\
\11/
·x1 +
/00\
\00/
·x2

POL(s(x1)) =
/1\
\0/
 +
/01\
\10/
·x1

POL(h(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(f(x1)) =
/0\
\1/
 +
/10\
\10/
·x1

POL(i(x1, x2)) =
/0\
\0/
 +
/10\
\00/
·x1 +
/00\
\00/
·x2

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

POL(c) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

f(g(s(x), y)) → f(g(x, s(y)))
i(x, x) → i(a, b)
f(s(x)) → s(s(f(h(s(x)))))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))

(51) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(s(x)) → F(h(s(x)))

The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
h(i(x, y)) → i(i(c, h(h(y))), x)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
i(x, x) → i(a, b)

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(s(x)) → F(h(s(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(F(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

POL(s(x1)) =
/0\
\1/
 +
/00\
\10/
·x1

POL(h(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(f(x1)) =
/1\
\0/
 +
/00\
\01/
·x1

POL(g(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(i(x1, x2)) =
/0\
\0/
 +
/11\
\11/
·x1 +
/00\
\00/
·x2

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

POL(c) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

f(g(s(x), y)) → f(g(x, s(y)))
i(x, x) → i(a, b)
f(s(x)) → s(s(f(h(s(x)))))
h(i(x, y)) → i(i(c, h(h(y))), x)
g(x, x) → g(a, b)
h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))

(53) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

h(s(f(x))) → h(f(x))
h(g(x, s(y))) → h(g(s(x), y))
g(x, x) → g(a, b)
h(i(x, y)) → i(i(c, h(h(y))), x)
f(s(x)) → s(s(f(h(s(x)))))
f(g(s(x), y)) → f(g(x, s(y)))
i(x, x) → i(a, b)

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

(54) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(55) TRUE