Termination w.r.t. Q proof of /tmp/tmpXe2JhS/#3.10.trs

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

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 DependencyGraphProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 PisEmptyProof (⇔)

    ↳31 TRUE

  ↳32 QDP

(0) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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:

EQ(s(x), s(y)) → EQ(x, y)
LE(s(x), s(y)) → LE(x, y)
APP(add(n, x), y) → APP(x, y)
MIN(add(n, add(m, x))) → IF_MIN(le(n, m), add(n, add(m, x)))
MIN(add(n, add(m, x))) → LE(n, m)
IF_MIN(true, add(n, add(m, x))) → MIN(add(n, x))
IF_MIN(false, add(n, add(m, x))) → MIN(add(m, x))
RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
RM(n, add(m, x)) → EQ(n, m)
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)
MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
MINSORT(add(n, x), y) → EQ(n, min(add(n, x)))
MINSORT(add(n, x), y) → MIN(add(n, x))
IF_MINSORT(true, add(n, x), y) → MINSORT(app(rm(n, x), y), nil)
IF_MINSORT(true, add(n, x), y) → APP(rm(n, x), y)
IF_MINSORT(true, add(n, x), y) → RM(n, x)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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 6 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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.

APP(add(n, x), y) → APP(x, y)

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

Lexicographic path order with status [LPO].
Precedence:

add₂ > APP₁

Status:

APP₁: [1]
add₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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.

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) = LE(x1)
s(x1) = s(x1)

Lexicographic path order with status [LPO].
Precedence:

s₁ > LE₁

Status:

s₁: [1]
LE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, x1), x2)

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:

MIN(add(n, add(m, x))) → IF_MIN(le(n, m), add(n, add(m, x)))
IF_MIN(true, add(n, add(m, x))) → MIN(add(n, x))
IF_MIN(false, add(n, add(m, x))) → MIN(add(m, x))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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.

IF_MIN(true, add(n, add(m, x))) → MIN(add(n, x))
IF_MIN(false, add(n, add(m, x))) → MIN(add(m, x))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1) = x1
add(x1, x2) = add(x2)
IF_MIN(x1, x2) = x2
le(x1, x2) = le
true = true
false = false
s(x1) = s
0 = 0

Lexicographic path order with status [LPO].
Precedence:

le > true
le > false > add₁
0 > true

Status:

add₁: [1]
true: []
false: []
s: []
0: []
le: []

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

MIN(add(n, add(m, x))) → IF_MIN(le(n, m), add(n, add(m, x)))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, x1), x2)

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

(20) DependencyGraphProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, 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.

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) = EQ(x1)
s(x1) = s(x1)

Lexicographic path order with status [LPO].
Precedence:

s₁ > EQ₁

Status:

EQ₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, x1), x2)

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, x1), x2)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
RM(x1, x2) = RM(x2)
add(x1, x2) = add(x1, x2)
IF_RM(x1, x2, x3) = x3
eq(x1, x2) = eq(x1)
true = true
false = false
s(x1) = x1
0 = 0

Lexicographic path order with status [LPO].
Precedence:

add₂ > eq₁ > false > RM₁
true > RM₁
0 > false > RM₁

Status:

add₂: [1,2]
eq₁: [1]
true: []
false: []
0: []
RM₁: [1]

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, x1), x2)

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

IF_MINSORT(true, add(n, x), y) → MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(add(n, nil)) → n
min(add(n, add(m, x))) → if_min(le(n, m), add(n, add(m, x)))
if_min(true, add(n, add(m, x))) → min(add(n, x))
if_min(false, add(n, add(m, x))) → min(add(m, x))
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
min(add(x0, nil))
min(add(x0, add(x1, x2)))
if_min(true, add(x0, add(x1, x2)))
if_min(false, add(x0, add(x1, x2)))
rm(x0, nil)
rm(x0, add(x1, x2))
if_rm(true, x0, add(x1, x2))
if_rm(false, x0, add(x1, x2))
minsort(nil, nil)
minsort(add(x0, x1), x2)
if_minsort(true, add(x0, x1), x2)
if_minsort(false, add(x0, x1), x2)

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