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

(0) Obligation:

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

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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

(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(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(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(maxlist, x), app(app(cons, y), ys)) → APP(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(if, app(app(le, x), y))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(le, x), y)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(le, x)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), ys)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(maxlist, y)
APP(height, app(app(node, x), xs)) → APP(s, app(app(maxlist, 0), app(app(map, height), xs)))
APP(height, app(app(node, x), xs)) → APP(app(maxlist, 0), app(app(map, height), xs))
APP(height, app(app(node, x), xs)) → APP(maxlist, 0)
APP(height, app(app(node, x), xs)) → APP(app(map, height), xs)
APP(height, app(app(node, x), xs)) → APP(map, height)

The TRS R consists of the following rules:

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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 12 less nodes.

(6) Complex Obligation (AND)

(7) 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(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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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

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)  =  le1(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
s1 > le11

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(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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), ys)

The TRS R consists of the following rules:

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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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

maxlist1(x, cons(y, ys)) → maxlist1(y, ys)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), ys)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
cons2 > maxlist12

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(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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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(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(height, app(app(node, x), xs)) → APP(app(map, height), xs)

The TRS R consists of the following rules:

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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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.


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

Recursive Path Order [RPO].
Precedence:
APP1 > app2 > map
cons > app2 > map
height > app2 > map
node > app2 > map

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(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(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(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(app(maxlist, x0), app(app(cons, x1), x2))
app(app(maxlist, x0), nil)
app(height, app(app(node, x0), x1))

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