(0) Obligation:

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

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))
APP(app(h, z), app(e, x)) → APP(h, app(c, z))
APP(app(h, z), app(e, x)) → APP(c, z)
APP(app(h, z), app(e, x)) → APP(app(d, z), x)
APP(app(h, z), app(e, x)) → APP(d, z)
APP(app(d, z), app(app(g, 0), 0)) → APP(e, 0)
APP(app(d, z), app(app(g, x), y)) → APP(app(g, app(e, x)), app(app(d, z), y))
APP(app(d, z), app(app(g, x), y)) → APP(g, app(e, x))
APP(app(d, z), app(app(g, x), y)) → APP(e, x)
APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(g, app(app(d, app(c, z)), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(d, z)
APP(app(g, app(e, x)), app(e, y)) → APP(e, app(app(g, x), y))
APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)
APP(app(g, app(e, x)), app(e, y)) → APP(g, x)
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(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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 4 SCCs with 23 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)

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

(8) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(9) Obligation:

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

g(e(x), e(y)) → g(x, y)

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

(10) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • g(e(x), e(y)) → g(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(11) TRUE

(12) Obligation:

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

APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(13) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

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

APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))

The TRS R consists of the following rules:

app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), y))

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
    The graph contains the following edges 2 > 2

(16) TRUE

(17) Obligation:

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

APP(app(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(18) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule APP(app(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x)) we obtained the following new rules [LPAR04]:

APP(app(h, app(c, y_0)), app(e, x1)) → APP(app(h, app(c, app(c, y_0))), app(app(d, app(c, y_0)), x1))

(19) Obligation:

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

APP(app(h, app(c, y_0)), app(e, x1)) → APP(app(h, app(c, app(c, y_0))), app(app(d, app(c, y_0)), x1))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(20) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule APP(app(h, app(c, y_0)), app(e, x1)) → APP(app(h, app(c, app(c, y_0))), app(app(d, app(c, y_0)), x1)) we obtained the following new rules [LPAR04]:

APP(app(h, app(c, app(c, y_0))), app(e, x1)) → APP(app(h, app(c, app(c, app(c, y_0)))), app(app(d, app(c, app(c, y_0))), x1))

(21) Obligation:

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

APP(app(h, app(c, app(c, y_0))), app(e, x1)) → APP(app(h, app(c, app(c, app(c, y_0)))), app(app(d, app(c, app(c, y_0))), x1))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(22) 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(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), 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))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(23) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(24) 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(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

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

(25) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
    The graph contains the following edges 2 >= 2

  • APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
    The graph contains the following edges 2 >= 2

(26) TRUE