(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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) 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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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)
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
(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(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) → APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) → APP(minus, 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(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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)
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
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 14 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
The TRS R consists of the following rules:
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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)
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
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, x), app(s, y)) → APP(app(minus, x), y)
R is empty.
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
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(x, s(y)) → minus1(x, y)
R is empty.
The set Q consists of the following terms:
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)
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].
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
minus1(x, s(y)) → minus1(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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:
- minus1(x, s(y)) → minus1(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, app(app(minus, x), y)), app(s, y))
The TRS R consists of the following rules:
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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)
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(quot, app(s, x)), app(s, y)) → APP(app(quot, 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, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(pred, app(s, x)) → x
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(19) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
quot1(s(x), s(y)) → quot1(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
The set Q consists of the following terms:
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)
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].
quot(0, s(x0))
quot(s(x0), s(x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
quot1(s(x), s(y)) → quot1(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
The set Q consists of the following terms:
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
We have to consider all minimal (P,Q,R)-chains.
(23) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
quot1(s(x), s(y)) → quot1(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(quot1(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(minus(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
minus(x, 0) → x
(24) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
The set Q consists of the following terms:
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
We have to consider all minimal (P,Q,R)-chains.
(25) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(26) TRUE
(27) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
quot1(s(x), s(y)) → quot1(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(minus(x1, x2)) = x1
POL(pred(x1)) = x1
POL(quot1(x1, x2)) = x1
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
minus(x, 0) → x
(28) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → x
The set Q consists of the following terms:
pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
We have to consider all minimal (P,Q,R)-chains.
(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)
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(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
The TRS R consists of the following rules:
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) → 0
app(app(quot, app(s, x)), app(s, y)) → app(s, app(app(quot, app(app(minus, x), y)), app(s, 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)
The set Q consists of the following terms:
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(quot, 0), app(s, x0))
app(app(quot, app(s, x0)), app(s, x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(30) 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(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
The graph contains the following edges 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
(31) TRUE