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

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

(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:

LE(s(x), s(y)) → LE(x, y)
EQ(s(x), s(y)) → EQ(x, y)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)
IF2(false, x, y, xs) → DEL(x, xs)
MINSORT(cons(x, y)) → MIN(x, y)
MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))
MINSORT(cons(x, y)) → DEL(min(x, y), cons(x, y))
MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
MIN(x, cons(y, z)) → LE(x, y)
DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
DEL(x, cons(y, z)) → EQ(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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.

EQ(s(x), s(y)) → EQ(x, y)

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

Recursive path order with status [RPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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:

DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)
IF2(false, x, y, xs) → DEL(x, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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.

DEL(x, cons(y, z)) → IF2(eq(x, y), x, y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DEL(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
IF2(x1, x2, x3, x4)  =  x4

Recursive path order with status [RPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

IF2(false, x, y, xs) → DEL(x, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(16) TRUE

(17) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

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

Recursive path order with status [RPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)
IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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.

MIN(x, cons(y, z)) → IF1(le(x, y), x, y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
IF1(x1, x2, x3, x4)  =  x4

Recursive path order with status [RPO].
Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

IF1(true, x, y, xs) → MIN(x, xs)
IF1(false, x, y, xs) → MIN(y, xs)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(26) TRUE

(27) Obligation:

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

MINSORT(cons(x, y)) → MINSORT(del(min(x, y), cons(x, y)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
minsort(nil)
minsort(cons(x0, x1))
min(x0, nil)
min(x0, cons(x1, x2))
del(x0, nil)
del(x0, cons(x1, x2))

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