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 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 ATransformationProof (⇔)
    ↳11 QDP
    ↳12 QReductionProof (⇔)
    ↳13 QDP
    ↳14 MRRProof (⇔)
    ↳15 QDP
    ↳16 DependencyGraphProof (⇔)
    ↳17 TRUE
  ↳18 QDP
    ↳19 UsableRulesProof (⇔)
    ↳20 QDP
    ↳21 ATransformationProof (⇔)
    ↳22 QDP
    ↳23 QReductionProof (⇔)
    ↳24 QDP
    ↳25 QDPOrderProof (⇔)
    ↳26 QDP
    ↳27 PisEmptyProof (⇔)
    ↳28 TRUE
  ↳29 QDP
    ↳30 UsableRulesProof (⇔)
    ↳31 QDP
    ↳32 QDPSizeChangeProof (⇔)
    ↳33 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, 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(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, app(p, app(s, x)))
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, x))
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, y))
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), y)), app(s, y)))
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))
APP(app(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), y))
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(div, app(s, x)), app(s, y)) → APP(minus, 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(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, 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 9 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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

(10) ATransformationProof (EQUIVALENT transformation)

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

(11) Obligation:

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

minus1(s(x), s(y)) → minus1(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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

(12) 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].

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))

(13) Obligation:

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

minus1(s(x), s(y)) → minus1(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

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

(14) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

p(s(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(minus1(x1, x2)) = x1 + x2   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   

(15) Obligation:

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

minus1(s(x), s(y)) → minus1(p(s(x)), p(s(y)))

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

p(s(x0))

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

(16) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(17) TRUE

(18) Obligation:

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

APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, 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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), y)), app(s, y))

The TRS R consists of the following rules:

app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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

(21) ATransformationProof (EQUIVALENT transformation)

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

(22) Obligation:

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

div1(s(x), s(y)) → div1(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(x)) → x

The set Q consists of the following terms:

map(x0, nil)
map(x0, cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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

(23) 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].

map(x0, nil)
map(x0, cons(x1, x2))
div(0, s(x0))
div(s(x0), s(x1))

(24) Obligation:

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

div1(s(x), s(y)) → div1(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(x)) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


div1(s(x), s(y)) → div1(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(div1(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

p(s(x)) → x
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, 0) → x

(26) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(x)) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) 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)

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(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), y)), app(s, y)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)

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

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(minus, x0), 0)
app(app(minus, app(s, x0)), app(s, x1))
app(p, app(s, x0))
app(app(div, 0), app(s, x0))
app(app(div, app(s, x0)), app(s, x1))

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

(32) 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, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
    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

(33) TRUE