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 Instantiation (⇔)
    ↳9 QDP
    ↳10 DependencyGraphProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 UsableRulesProof (⇔)
    ↳14 QDP
    ↳15 Narrowing (⇔)
    ↳16 QDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 UsableRulesProof (⇔)
    ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 PisEmptyProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 UsableRulesProof (⇔)
    ↳32 QDP
    ↳33 Instantiation (⇔)
    ↳34 QDP
    ↳35 Instantiation (⇔)
    ↳36 QDP
    ↳37 Instantiation (⇔)
    ↳38 QDP
    ↳39 DependencyGraphProof (⇔)
    ↳40 QDP
    ↳41 Narrowing (⇔)
    ↳42 QDP
    ↳43 ForwardInstantiation (⇔)
    ↳44 QDP
    ↳45 ForwardInstantiation (⇔)
    ↳46 QDP
    ↳47 DependencyGraphProof (⇔)
    ↳48 TRUE

(0) Obligation:

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

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

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:

G(x, x, x) → G(c, d, e)
G(x, y, x) → G(c, d, e)
S(f(x, y)) → F(y, f(s(s(x)), a))
S(f(x, y)) → F(s(s(x)), a)
S(f(x, y)) → S(s(x))
S(f(x, y)) → S(x)
H(h(x, a), y) → H(h(a, y), h(a, x))
H(h(x, a), y) → H(a, y)
H(h(x, a), y) → H(a, x)
F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))
F(x, f(y, f(x, y))) → F(x, f(y, b))
F(x, f(y, f(x, y))) → F(y, b)
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(h(a, y), g(x, b, a)) → F(x, s(y))
F(h(a, y), g(x, b, a)) → S(y)
F(h(a, y), g(x, b, a)) → S(b)
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
H(f(x, s(y)), b) → G(y, a, f(s(x), a))
H(f(x, s(y)), b) → F(s(x), a)
H(f(x, s(y)), b) → S(x)
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(x, g(x, a, f(s(x), y))) → G(a, b, y)

The TRS R consists of the following rules:

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

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 4 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))

The TRS R consists of the following rules:

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

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:

F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))

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

(8) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, f(y, f(x, y))) → F(a, f(x, f(y, b))) we obtained the following new rules [LPAR04]:

F(a, f(b, f(a, b))) → F(a, f(a, f(b, b)))

(9) Obligation:

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

F(a, f(b, f(a, b))) → F(a, f(a, f(b, b)))

R is empty.
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 0 SCCs with 1 less node.

(11) TRUE

(12) Obligation:

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

S(f(x, y)) → S(x)
S(f(x, y)) → S(s(x))

The TRS R consists of the following rules:

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

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

(13) 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.

(14) Obligation:

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

S(f(x, y)) → S(x)
S(f(x, y)) → S(s(x))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(15) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule S(f(x, y)) → S(s(x)) at position [0] we obtained the following new rules [LPAR04]:

S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))
S(f(x0, y1)) → S(b)

(16) Obligation:

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

S(f(x, y)) → S(x)
S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))
S(f(x0, y1)) → S(b)

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

S(f(x, y)) → S(x)
S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))

The TRS R consists of the following rules:

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(f(x, y)) → S(x)
S(f(f(x0, x1), y1)) → S(f(x1, f(s(s(x0)), a)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(S(x1)) =
/0\
\0/
 +
/20\
\00/
·x1

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

POL(s(x1)) =
/0\
\0/
 +
/01\
\30/
·x1

POL(a) =
/0\
\0/

POL(b) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

s(y) → b
s(f(x, y)) → f(y, f(s(s(x)), a))

(20) Obligation:

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

s(f(x, y)) → f(y, f(s(s(x)), a))
s(y) → b

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

H(h(x, a), y) → H(h(a, y), h(a, x))

The TRS R consists of the following rules:

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

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:

H(h(x, a), y) → H(h(a, y), h(a, x))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(H(x1, x2)) =
/1\
\0/
 +
/01\
\00/
·x1 +
/10\
\00/
·x2

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

POL(a) =
/1\
\0/

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)

The TRS R consists of the following rules:

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

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(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)

The TRS R consists of the following rules:

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

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y)) we obtained the following new rules [LPAR04]:

F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))

(34) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))

The TRS R consists of the following rules:

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

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

(35) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b)) we obtained the following new rules [LPAR04]:

F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))

(36) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))

The TRS R consists of the following rules:

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

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

(37) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, g(x, a, f(s(x), y))) → H(x, b) we obtained the following new rules [LPAR04]:

F(a, g(a, a, f(s(a), x1))) → H(a, b)

(38) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))
F(a, g(a, a, f(s(a), x1))) → H(a, b)

The TRS R consists of the following rules:

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

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) Obligation:

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

F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b))
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))

The TRS R consists of the following rules:

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

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

(41) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(h(a, b), g(x1, b, a)) → H(f(x1, s(b)), s(b)) at position [1] we obtained the following new rules [LPAR04]:

F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)

(42) Obligation:

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

F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1))
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)

The TRS R consists of the following rules:

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

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

(43) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(a, g(a, a, f(s(a), x1))) → F(h(a, b), g(a, b, x1)) we obtained the following new rules [LPAR04]:

F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a))

(44) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)
F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a))

The TRS R consists of the following rules:

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

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

(45) ForwardInstantiation (EQUIVALENT transformation)

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

H(f(a, s(a)), b) → F(a, g(a, a, f(s(a), a)))

(46) Obligation:

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

F(h(a, b), g(y0, b, a)) → H(f(y0, s(b)), b)
F(a, g(a, a, f(s(a), a))) → F(h(a, b), g(a, b, a))
H(f(a, s(a)), b) → F(a, g(a, a, f(s(a), a)))

The TRS R consists of the following rules:

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

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

(47) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(48) TRUE