Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 MRRProof (⇔)
    ↳9 QDP
    ↳10 DependencyGraphProof (⇔)
    ↳11 QDP
    ↳12 MRRProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 UsableRulesProof (⇔)
      ↳18 QDP
      ↳19 Narrowing (⇔)
      ↳20 QDP
      ↳21 DependencyGraphProof (⇔)
      ↳22 QDP
      ↳23 QDPOrderProof (⇔)
      ↳24 QDP
    ↳25 UsableRulesProof (⇔)
      ↳26 QDP

(0) Obligation:

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

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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:

F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))
F(x, f(s(s(y)), nil)) → F(s(0), nil)

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2)) = x1 + x2   
POL(f(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(s(x1)) = 1 + x1   

(9) Obligation:

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

F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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 1 less node.

(11) Obligation:

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

F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(12) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2)) = 2·x1 + 2·x2   
POL(f(x1, x2)) = 2·x1 + x2   
POL(nil) = 0   
POL(s(x1)) = 2 + x1   

(13) Obligation:

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

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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

(17) 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.

(18) Obligation:

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

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w)))) at position [0] we obtained the following new rules [LPAR04]:

L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(x0)), f(s(x1), nil))) → L1(f(s(0), f(s(x0), f(x1, f(s(0), nil)))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))

(20) Obligation:

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

L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(x0)), f(s(x1), nil))) → L1(f(s(0), f(s(x0), f(x1, f(s(0), nil)))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(L1(x1)) = (2)x1   
POL(f(x1, x2)) = (1/2)x2   
POL(s(x1)) = (1/4)x1   
POL(0) = 0   
POL(nil) = 4   
The value of delta used in the strict ordering is 3/4.
The following usable rules [FROCOS05] were oriented:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

(24) Obligation:

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

L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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

(25) 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.

(26) Obligation:

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

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

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