Termination w.r.t. Q of the given QTRS could be proven:

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 QDPOrderProof (⇔)
    ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE

(0) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), 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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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:

APP(app(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)
APP(app(sub, app(s, x)), app(s, y)) → APP(sub, x)
APP(app(gtr, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(gtr, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(app(d, app(s, x)), app(s, y)) → APP(app(if, app(app(gtr, x), y)), false)
APP(app(d, app(s, x)), app(s, y)) → APP(if, app(app(gtr, x), y))
APP(app(d, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(d, app(s, x)), app(s, y)) → APP(app(sub, y), x)
APP(app(d, app(s, x)), app(s, y)) → APP(sub, y)
APP(len, app(app(cons, x), xs)) → APP(s, app(len, xs))
APP(len, app(app(cons, x), xs)) → APP(len, xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))
APP(app(filter, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, p), xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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 14 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(len, app(app(cons, x), xs)) → APP(len, xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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

len1(cons(x, xs)) → len1(xs)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(len, app(app(cons, x), xs)) → APP(len, xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
len1(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
cons2: [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:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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:

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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

gtr1(s(x), s(y)) → gtr1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(gtr, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
gtr1(x1, x2)  =  gtr1(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
s1 > gtr11

Status:
s1: [1]
gtr11: [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:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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

sub1(s(x), s(y)) → sub1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
sub1(x1, x2)  =  sub1(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
s1 > sub11

Status:
sub11: [1]
s1: [1]

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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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:

APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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

d1(s(x), s(y)) → d1(s(x), sub(y, x))

The a-transformed usable rules are

sub(x, 0) → x
sub(s(x), s(y)) → sub(x, y)


The following pairs can be oriented strictly and are deleted.


APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
d1(x1, x2)  =  x2
s(x1)  =  s(x1)
sub(x1, x2)  =  sub(x1)
0  =  0

Lexicographic path order with status [LPO].
Precedence:
s1 > sub1

Status:
sub1: [1]
s1: [1]
0: []

The following usable rules [FROCOS05] were oriented:

app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)

(24) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, 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.


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

Lexicographic path order with status [LPO].
Precedence:
APP1 > app1
filter > app1

Status:
APP1: [1]
cons: []
app1: [1]
filter: []

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(30) 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

filter1(p, cons(x, xs)) → filter1(p, xs)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
cons2 > filter12

Status:
cons2: [1,2]
filter12: [1,2]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE