YES Termination w.r.t. Q proof of AProVE_07_kabasci03.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 76 ms)
↳8 QDP
↳9 QDPSizeChangeProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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:

H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
H(c(x, y), c(s(z), z), t(w)) → T(t(c(x, c(y, t(w)))))
H(c(x, y), c(s(z), z), t(w)) → T(c(x, c(y, t(w))))
H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
H(x, c(y, z), t(w)) → T(c(t(w), w))
H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
H(c(s(x), c(s(0), y)), z, t(x)) → T(t(c(x, s(x))))
H(c(s(x), c(s(0), y)), z, t(x)) → T(c(x, s(x)))
T(t(x)) → T(c(t(x), x))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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 1 SCC with 6 less nodes.

(4) Obligation:

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

H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


H(c(x, y), c(s(z), z), t(w)) → H(z, c(y, x), t(t(c(x, c(y, t(w))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(H(x1, x2, x3)) = x1 + x2   
POL(c(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = x1   
POL(t(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(6) Obligation:

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

H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


H(c(s(x), c(s(0), y)), z, t(x)) → H(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = [2]   
POL(H(x1, x2, x3)) = [4]x1 + [2]x2   
POL(c(x1, x2)) = [4]x1 + x2   
POL(s(x1)) = [1/2]x1   
POL(t(x1)) = [4]   
The value of delta used in the strict ordering is 8.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(8) Obligation:

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

H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))

The TRS R consists of the following rules:

h(c(x, y), c(s(z), z), t(w)) → h(z, c(y, x), t(t(c(x, c(y, t(w))))))
h(x, c(y, z), t(w)) → h(c(s(y), x), z, t(c(t(w), w)))
h(c(s(x), c(s(0), y)), z, t(x)) → h(y, c(s(0), c(x, z)), t(t(c(x, s(x)))))
t(t(x)) → t(c(t(x), x))
t(x) → x
t(x) → c(0, c(0, c(0, c(0, c(0, x)))))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

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:

  • H(x, c(y, z), t(w)) → H(c(s(y), x), z, t(c(t(w), w)))
    The graph contains the following edges 2 > 2

(10) YES