Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DOUBLE → APP(s, app(s, 0))
APP(app(times, app(s, x)), y) → APP(times, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
INC → APP(map, app(plus, app(s, 0)))
DOUBLE → APP(s, 0)
DOUBLE → APP(times, app(s, app(s, 0)))
APP(app(times, app(s, x)), y) → APP(app(plus, app(app(times, x), y)), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
DOUBLE → APP(map, app(times, app(s, app(s, 0))))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
INC → APP(plus, app(s, 0))
APP(app(plus, app(s, x)), y) → APP(plus, x)
INC → APP(s, 0)
APP(app(times, app(s, x)), y) → APP(app(times, x), y)
APP(app(times, app(s, x)), y) → APP(plus, app(app(times, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

DOUBLE → APP(s, app(s, 0))
APP(app(times, app(s, x)), y) → APP(times, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
INC → APP(map, app(plus, app(s, 0)))
DOUBLE → APP(s, 0)
DOUBLE → APP(times, app(s, app(s, 0)))
APP(app(times, app(s, x)), y) → APP(app(plus, app(app(times, x), y)), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
DOUBLE → APP(map, app(times, app(s, app(s, 0))))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
INC → APP(plus, app(s, 0))
APP(app(plus, app(s, x)), y) → APP(plus, x)
INC → APP(s, 0)
APP(app(times, app(s, x)), y) → APP(app(times, x), y)
APP(app(times, app(s, x)), y) → APP(plus, app(app(times, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

DOUBLE → APP(s, app(s, 0))
APP(app(times, app(s, x)), y) → APP(times, x)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
INC → APP(map, app(plus, app(s, 0)))
DOUBLE → APP(times, app(s, app(s, 0)))
DOUBLE → APP(s, 0)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(times, app(s, x)), y) → APP(app(plus, app(app(times, x), y)), y)
DOUBLE → APP(map, app(times, app(s, app(s, 0))))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
INC → APP(plus, app(s, 0))
APP(app(plus, app(s, x)), y) → APP(plus, x)
INC → APP(s, 0)
APP(app(times, app(s, x)), y) → APP(app(times, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(times, app(s, x)), y) → APP(plus, app(app(times, x), y))

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 14 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

PLUS(s(x), y) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
map(x0, nil)
map(x0, cons(x1, x2))
inc
double

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


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.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

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

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

TIMES(s(x), y) → TIMES(x, y)

R is empty.
The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
map(x0, nil)
map(x0, cons(x1, x2))
inc
double

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


The following pairs can be oriented strictly and are deleted.


APP(app(times, app(s, x)), y) → APP(app(times, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

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)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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)
The remaining pairs can at least be oriented weakly.
none
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 [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
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))
inc → app(map, app(plus, app(s, 0)))
double → app(map, app(times, app(s, app(s, 0))))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.