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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

(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)
F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
F(s(x), s(y), s(z)) → MAX(s(x), max(s(y), s(z)))
F(s(x), s(y), s(z)) → MAX(s(y), s(z))
F(s(x), s(y), s(z)) → P(min(s(x), max(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(x), max(s(y), s(z)))
F(s(x), s(y), s(z)) → MIN(s(x), min(s(y), s(z)))
F(s(x), s(y), s(z)) → MIN(s(y), s(z))
F(0, y, z) → MAX(y, z)
F(x, 0, z) → MAX(x, z)
F(x, y, 0) → 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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 9 less nodes.

(6) Complex Obligation (AND)

(7) 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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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.


MAX(s(x), s(y)) → MAX(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MAX(x1, x2)  =  MAX(x1)
s(x1)  =  s(x1)

Recursive path order with status [RPO].
Precedence:
trivial

Status:
MAX1: [1]
s1: multiset

The following usable rules [FROCOS05] were oriented: none

(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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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:

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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MIN(s(x), s(y)) → MIN(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2)  =  MIN(x1)
s(x1)  =  s(x1)

Recursive path order with status [RPO].
Precedence:
trivial

Status:
MIN1: [1]
s1: multiset

The following usable rules [FROCOS05] were oriented: none

(14) 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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

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))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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))
p(s(x0))
f(s(x0), s(x1), s(x2))
f(0, x0, x1)
f(x0, 0, x1)
f(x0, x1, 0)

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