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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) 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(sum, app(app(cons, x), xs)) → APP(app(plus, x), app(sum, xs))
APP(sum, app(app(cons, x), xs)) → APP(plus, x)
APP(sum, app(app(cons, x), xs)) → APP(sum, xs)
APP(size, app(app(node, x), xs)) → APP(s, app(sum, app(app(map, size), xs)))
APP(size, app(app(node, x), xs)) → APP(sum, app(app(map, size), xs))
APP(size, app(app(node, x), xs)) → APP(app(map, size), xs)
APP(size, app(app(node, x), xs)) → APP(map, size)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, x)

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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(plus, app(s, x)), y) → APP(app(plus, 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

APP(sum, app(app(cons, x), xs)) → APP(sum, 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(12) 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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(size, app(app(node, x), xs)) → APP(app(map, size), 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(16) 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(size, app(app(node, x), xs)) → APP(app(map, size), xs)
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)
map  =  map
cons  =  cons
size  =  size
node  =  node

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(17) 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(sum, app(app(cons, x), xs)) → app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) → app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE