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:

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) 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.

(9) 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.

(10) 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.

(11) 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.