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

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
  ↳13 QDP
  ↳14 QDP
  ↳15 QDP
  ↳16 QDP

(0) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
min(0, y) → 0

The TRS R 2 is

f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The signature Sigma is {f}

(2) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
+1(s(x), y) → +1(x, y)
-1(s(x), s(y)) → -1(x, y)
*1(x, s(y)) → +1(x, *(x, y))
*1(x, s(y)) → *1(x, y)
F(s(x)) → F(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))
F(s(x)) → -1(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0)))))))
F(s(x)) → MAX(*(s(x), s(x)), +(s(x), s(s(s(0)))))
F(s(x)) → *1(s(x), s(x))
F(s(x)) → +1(s(x), s(s(s(0))))
F(s(x)) → MAX(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))
F(s(x)) → +1(s(x), s(s(s(s(0)))))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


-1(s(x), s(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
s(x1)  =  s(x1)
min(x1, x2)  =  x2
0  =  0
max(x1, x2)  =  max(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  x1
*(x1, x2)  =  *(x1, x2)
f(x1)  =  f

Lexicographic path order with status [LPO].
Quasi-Precedence:
max2 > s1 > 0
*2 > +2 > s1 > 0
f > 0

Status:
trivial


The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

(9) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(13) Obligation:

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

*1(x, s(y)) → *1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(14) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(15) Obligation:

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

F(s(x)) → F(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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

(16) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
f(s(x)) → f(-(max(*(s(x), s(x)), +(s(x), s(s(s(0))))), max(s(*(s(x), s(x))), +(s(x), s(s(s(s(0))))))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
f(s(x0))

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