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 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 TRUE
  ↳27 QDP
  ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 DependencyGraphProof (⇔)
    ↳32 TRUE

(0) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), 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:

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)
APP(app(eq, app(s, x)), app(s, y)) → APP(eq, x)
APP(minsort, app(app(cons, x), y)) → APP(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
APP(minsort, app(app(cons, x), y)) → APP(cons, app(app(min, x), y))
APP(minsort, app(app(cons, x), y)) → APP(app(min, x), y)
APP(minsort, app(app(cons, x), y)) → APP(min, x)
APP(minsort, app(app(cons, x), y)) → APP(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y)))
APP(minsort, app(app(cons, x), y)) → APP(app(del, app(app(min, x), y)), app(app(cons, x), y))
APP(minsort, app(app(cons, x), y)) → APP(del, app(app(min, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
APP(app(min, x), app(app(cons, y), z)) → APP(app(if, app(app(le, x), y)), app(app(min, x), z))
APP(app(min, x), app(app(cons, y), z)) → APP(if, app(app(le, x), y))
APP(app(min, x), app(app(cons, y), z)) → APP(app(le, x), y)
APP(app(min, x), app(app(cons, y), z)) → APP(le, x)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)
APP(app(min, x), app(app(cons, y), z)) → APP(min, y)
APP(app(del, x), app(app(cons, y), z)) → APP(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
APP(app(del, x), app(app(cons, y), z)) → APP(app(if, app(app(eq, x), y)), z)
APP(app(del, x), app(app(cons, y), z)) → APP(if, app(app(eq, x), y))
APP(app(del, x), app(app(cons, y), z)) → APP(app(eq, x), y)
APP(app(del, x), app(app(cons, y), z)) → APP(eq, x)
APP(app(del, x), app(app(cons, y), z)) → APP(app(cons, y), app(app(del, x), z))
APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), 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 29 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

eq1(s(x), s(y)) → eq1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(eq, app(s, x)), app(s, y)) → APP(app(eq, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
eq1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
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:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), 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:

APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

del1(x, cons(y, z)) → del1(x, z)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(del, x), app(app(cons, y), z)) → APP(app(del, x), z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
del1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), 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:

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

le1(s(x), s(y)) → le1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
le1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
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:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), 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:

APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

min1(x, cons(y, z)) → min1(y, z)
min1(x, cons(y, z)) → min1(x, z)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(min, x), app(app(cons, y), z)) → APP(app(min, y), z)
APP(app(min, x), app(app(cons, y), z)) → APP(app(min, x), z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
min1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), 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:

APP(minsort, app(app(cons, x), y)) → APP(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y)))

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(28) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
cons  =  cons

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(eq, 0), 0) → true
app(app(eq, 0), app(s, y)) → false
app(app(eq, app(s, x)), 0) → false
app(app(eq, app(s, x)), app(s, y)) → app(app(eq, x), y)
app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(minsort, nil) → nil
app(minsort, app(app(cons, x), y)) → app(app(cons, app(app(min, x), y)), app(minsort, app(app(del, app(app(min, x), y)), app(app(cons, x), y))))
app(app(min, x), nil) → x
app(app(min, x), app(app(cons, y), z)) → app(app(app(if, app(app(le, x), y)), app(app(min, x), z)), app(app(min, y), z))
app(app(del, x), nil) → nil
app(app(del, x), app(app(cons, y), z)) → app(app(app(if, app(app(eq, x), y)), z), app(app(cons, y), app(app(del, x), z)))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(eq, 0), 0)
app(app(eq, 0), app(s, x0))
app(app(eq, app(s, x0)), 0)
app(app(eq, app(s, x0)), app(s, x1))
app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(minsort, nil)
app(minsort, app(app(cons, x0), x1))
app(app(min, x0), nil)
app(app(min, x0), app(app(cons, x1), x2))
app(app(del, x0), nil)
app(app(del, x0), app(app(cons, x1), x2))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) TRUE