YES Termination w.r.t. Q proof of AotoYamada_05_010.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 3 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 ATransformationProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QReductionProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 ATransformationProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QReductionProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QReductionProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 TransformationProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 TransformationProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 TransformationProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QDPOrderProof (⇔, 760 ms)
        ↳41 QDP
        ↳42 UsableRulesProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 QDPSizeChangeProof (⇔, 0 ms)
        ↳45 YES

(0) Obligation:

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

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(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(app(curry, g), x), y) → 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))
inc → app(map, app(app(curry, plus), app(s, 0)))
double → app(map, app(app(curry, 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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
inc
double

(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(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)
APP(app(times, app(s, x)), y) → APP(app(plus, app(app(times, x), y)), y)
APP(app(times, app(s, x)), y) → APP(plus, app(app(times, x), y))
APP(app(times, app(s, x)), y) → APP(app(times, x), y)
APP(app(times, app(s, x)), y) → APP(times, x)
APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(app(curry, g), x), 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)
INC → APP(map, app(app(curry, plus), app(s, 0)))
INC → APP(app(curry, plus), app(s, 0))
INC → APP(curry, plus)
INC → APP(s, 0)
DOUBLE → APP(map, app(app(curry, times), app(s, app(s, 0))))
DOUBLE → APP(app(curry, times), app(s, app(s, 0)))
DOUBLE → APP(curry, times)
DOUBLE → APP(s, app(s, 0))
DOUBLE → APP(s, 0)

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(app(curry, g), x), y) → 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))
inc → app(map, app(app(curry, plus), app(s, 0)))
double → app(map, app(app(curry, 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(app(curry, x0), x1), x2)
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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 16 less nodes.

(6) Complex Obligation (AND)

(7) 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(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(app(curry, g), x), y) → 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))
inc → app(map, app(app(curry, plus), app(s, 0)))
double → app(map, app(app(curry, 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(app(curry, x0), x1), x2)
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.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

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

R is empty.
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(app(curry, x0), x1), x2)
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.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

inc
double

(11) Obligation:

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

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

R is empty.
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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(12) ATransformationProof (EQUIVALENT transformation)

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

(13) Obligation:

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

plus1(s(x), y) → plus1(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)
curry(x0, x1, x2)
map(x0, nil)
map(x0, cons(x1, x2))

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

(14) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(15) Obligation:

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

plus1(s(x), y) → plus1(x, y)

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

(16) 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:

  • plus1(s(x), y) → plus1(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(17) YES

(18) Obligation:

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(app(curry, g), x), y) → 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))
inc → app(map, app(app(curry, plus), app(s, 0)))
double → app(map, app(app(curry, 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(app(curry, x0), x1), x2)
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.

(19) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

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

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

R is empty.
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(app(curry, x0), x1), x2)
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.

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

inc
double

(22) Obligation:

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

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

R is empty.
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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(23) ATransformationProof (EQUIVALENT transformation)

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

(24) Obligation:

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

times1(s(x), y) → times1(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)
curry(x0, x1, x2)
map(x0, nil)
map(x0, cons(x1, x2))

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

(25) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(26) Obligation:

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

times1(s(x), y) → times1(x, y)

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

(27) 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:

  • times1(s(x), y) → times1(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(28) YES

(29) Obligation:

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

APP(app(app(curry, g), x), y) → APP(g, x)
APP(app(app(curry, g), x), y) → APP(app(g, x), y)
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(app(curry, g), x), y) → 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))
inc → app(map, app(app(curry, plus), app(s, 0)))
double → app(map, app(app(curry, 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(app(curry, x0), x1), x2)
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.

(30) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(31) Obligation:

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

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

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(app(curry, x0), x1), x2)
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.

(32) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

inc
double

(33) Obligation:

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

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

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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(34) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule APP(app(app(curry, g), x), y) → APP(g, x) we obtained the following new rules [LPAR04]:

APP(app(app(curry, app(app(curry, y_0), y_1)), x1), x2) → APP(app(app(curry, y_0), y_1), x1) → APP(app(app(curry, app(app(curry, y_0), y_1)), x1), x2) → APP(app(app(curry, y_0), y_1), x1)
APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x2) → APP(app(map, y_0), app(app(cons, y_1), y_2)) → APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x2) → APP(app(map, y_0), app(app(cons, y_1), y_2))

(35) Obligation:

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

APP(app(app(curry, g), x), y) → APP(app(g, x), y)
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(app(curry, app(app(curry, y_0), y_1)), x1), x2) → APP(app(app(curry, y_0), y_1), x1)
APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x2) → APP(app(map, y_0), app(app(cons, y_1), y_2))

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(app(curry, g), x), y) → 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))

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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(36) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule APP(app(map, f), app(app(cons, x), xs)) → APP(f, x) we obtained the following new rules [LPAR04]:

APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), x2)) → APP(app(app(curry, y_0), y_1), x1) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), x2)) → APP(app(app(curry, y_0), y_1), x1)
APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), x2)) → APP(app(map, y_0), app(app(cons, y_1), y_2)) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), x2)) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), x2)) → APP(app(app(curry, app(app(curry, y_0), y_1)), y_2), x1) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), x2)) → APP(app(app(curry, app(app(curry, y_0), y_1)), y_2), x1)
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), x2)) → APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x1) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), x2)) → APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x1)

(37) Obligation:

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

APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(curry, app(app(curry, y_0), y_1)), x1), x2) → APP(app(app(curry, y_0), y_1), x1)
APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x2) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), x2)) → APP(app(app(curry, y_0), y_1), x1)
APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), x2)) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), x2)) → APP(app(app(curry, app(app(curry, y_0), y_1)), y_2), x1)
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), x2)) → APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x1)

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(app(curry, g), x), y) → 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))

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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(38) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs) we obtained the following new rules [LPAR04]:

APP(app(map, x0), app(app(cons, x1), app(app(cons, y_1), y_2))) → APP(app(map, x0), app(app(cons, y_1), y_2)) → APP(app(map, x0), app(app(cons, x1), app(app(cons, y_1), y_2))) → APP(app(map, x0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), app(app(cons, y_2), y_3))) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, y_2), y_3)) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), app(app(cons, y_2), y_3))) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, y_2), y_3))
APP(app(map, app(map, y_0)), app(app(cons, x1), app(app(cons, app(app(cons, y_1), y_2)), y_3))) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), y_3)) → APP(app(map, app(map, y_0)), app(app(cons, x1), app(app(cons, app(app(cons, y_1), y_2)), y_3))) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), y_3))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, y_3), y_4)) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, y_3), y_4))
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, y_3), y_4)) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, y_3), y_4))

(39) Obligation:

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

APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(app(curry, app(app(curry, y_0), y_1)), x1), x2) → APP(app(app(curry, y_0), y_1), x1)
APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x2) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), x2)) → APP(app(app(curry, y_0), y_1), x1)
APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), x2)) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), x2)) → APP(app(app(curry, app(app(curry, y_0), y_1)), y_2), x1)
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), x2)) → APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x1)
APP(app(map, x0), app(app(cons, x1), app(app(cons, y_1), y_2))) → APP(app(map, x0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), app(app(cons, y_2), y_3))) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, y_2), y_3))
APP(app(map, app(map, y_0)), app(app(cons, x1), app(app(cons, app(app(cons, y_1), y_2)), y_3))) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), y_3))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, y_3), y_4))
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, y_3), y_4))

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(app(curry, g), x), y) → 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))

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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


APP(app(app(curry, g), x), y) → APP(app(g, x), y)
APP(app(app(curry, app(app(curry, y_0), y_1)), x1), x2) → APP(app(app(curry, y_0), y_1), x1)
APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x2) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), x2)) → APP(app(app(curry, y_0), y_1), x1)
APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), x2)) → APP(app(map, y_0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), x2)) → APP(app(app(curry, app(app(curry, y_0), y_1)), y_2), x1)
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), x2)) → APP(app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2)), x1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(APP(x1, x2)) = x1   
POL(app(x1, x2)) = 1 + x1 + x1·x2   
POL(cons) = 0   
POL(curry) = 1   
POL(map) = 1   
POL(nil) = 0   
POL(plus) = 0   
POL(s) = 0   
POL(times) = 0   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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(app(curry, g), x), y) → 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))

(41) Obligation:

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

APP(app(map, x0), app(app(cons, x1), app(app(cons, y_1), y_2))) → APP(app(map, x0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), app(app(cons, y_2), y_3))) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, y_2), y_3))
APP(app(map, app(map, y_0)), app(app(cons, x1), app(app(cons, app(app(cons, y_1), y_2)), y_3))) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), y_3))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, y_3), y_4))
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, y_3), y_4))

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(app(curry, g), x), y) → 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))

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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(42) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(43) Obligation:

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

APP(app(map, x0), app(app(cons, x1), app(app(cons, y_1), y_2))) → APP(app(map, x0), app(app(cons, y_1), y_2))
APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), app(app(cons, y_2), y_3))) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, y_2), y_3))
APP(app(map, app(map, y_0)), app(app(cons, x1), app(app(cons, app(app(cons, y_1), y_2)), y_3))) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), y_3))
APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, y_3), y_4))
APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, y_3), y_4))

R is empty.
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(app(curry, x0), x1), x2)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(44) 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(map, app(map, y_0)), app(app(cons, x1), app(app(cons, app(app(cons, y_1), y_2)), y_3))) → APP(app(map, app(map, y_0)), app(app(cons, app(app(cons, y_1), y_2)), y_3))
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(map, x0), app(app(cons, x1), app(app(cons, y_1), y_2))) → APP(app(map, x0), app(app(cons, y_1), y_2))
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(map, y_0)), app(app(cons, y_1), y_2))), app(app(cons, y_3), y_4))
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, x1), app(app(cons, y_2), y_3))) → APP(app(map, app(app(curry, y_0), y_1)), app(app(cons, y_2), y_3))
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, x1), app(app(cons, y_3), y_4))) → APP(app(map, app(app(curry, app(app(curry, y_0), y_1)), y_2)), app(app(cons, y_3), y_4))
    The graph contains the following edges 1 >= 1, 2 > 2

(45) YES