(0) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

(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(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)
APP(app(lt, app(s, x)), app(s, y)) → APP(lt, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys))))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(lt, x), y))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(lt, x), y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(lt, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(merge, xs)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(eq, x), y))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(eq, x), y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(eq, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)
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(mult, app(s, x)), y) → APP(app(plus, y), app(app(mult, x), y))
APP(app(mult, app(s, x)), y) → APP(plus, y)
APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)
APP(app(mult, app(s, x)), y) → APP(mult, x)
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)
LIST1APP(app(map, app(mult, app(s, app(s, 0)))), hamming)
LIST1APP(map, app(mult, app(s, app(s, 0))))
LIST1APP(mult, app(s, app(s, 0)))
LIST1APP(s, app(s, 0))
LIST1APP(s, 0)
LIST1HAMMING
LIST2APP(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
LIST2APP(map, app(mult, app(s, app(s, app(s, 0)))))
LIST2APP(mult, app(s, app(s, app(s, 0))))
LIST2APP(s, app(s, app(s, 0)))
LIST2APP(s, app(s, 0))
LIST2APP(s, 0)
LIST2HAMMING
LIST3APP(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
LIST3APP(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0)))))))
LIST3APP(mult, app(s, app(s, app(s, app(s, app(s, 0))))))
LIST3APP(s, app(s, app(s, app(s, app(s, 0)))))
LIST3APP(s, app(s, app(s, app(s, 0))))
LIST3APP(s, app(s, app(s, 0)))
LIST3APP(s, app(s, 0))
LIST3APP(s, 0)
LIST3HAMMING
HAMMINGAPP(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))
HAMMINGAPP(cons, app(s, 0))
HAMMINGAPP(s, 0)
HAMMINGAPP(app(merge, list1), app(app(merge, list2), list3))
HAMMINGAPP(merge, list1)
HAMMINGLIST1
HAMMINGAPP(app(merge, list2), list3)
HAMMINGAPP(merge, list2)
HAMMINGLIST2
HAMMINGLIST3

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 48 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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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].

list1
list2
list3
hamming

(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(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

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:

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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].

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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].

list1
list2
list3
hamming

(22) Obligation:

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

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

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

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:

mult1(s(x), y) → mult1(x, y)

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

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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].

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

(26) Obligation:

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

mult1(s(x), y) → mult1(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:

  • mult1(s(x), y) → mult1(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(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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].

list1
list2
list3
hamming

(33) Obligation:

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

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

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

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

(34) ATransformationProof (EQUIVALENT transformation)

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

(35) Obligation:

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

lt1(s(x), s(y)) → lt1(x, y)

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

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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

(36) 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].

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

(37) Obligation:

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

lt1(s(x), s(y)) → lt1(x, y)

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

(38) 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:

  • lt1(s(x), s(y)) → lt1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(39) YES

(40) Obligation:

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

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(41) 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.

(42) Obligation:

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

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(43) 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].

list1
list2
list3
hamming

(44) Obligation:

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

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

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

(45) ATransformationProof (EQUIVALENT transformation)

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

(46) Obligation:

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

merge1(cons(x, xs), cons(y, ys)) → merge1(xs, ys)
merge1(cons(x, xs), cons(y, ys)) → merge1(xs, cons(y, ys))
merge1(cons(x, xs), cons(y, ys)) → merge1(cons(x, xs), ys)

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

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

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

(47) 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].

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

(48) Obligation:

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

merge1(cons(x, xs), cons(y, ys)) → merge1(xs, ys)
merge1(cons(x, xs), cons(y, ys)) → merge1(xs, cons(y, ys))
merge1(cons(x, xs), cons(y, ys)) → merge1(cons(x, xs), ys)

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

(49) 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:

  • merge1(cons(x, xs), cons(y, ys)) → merge1(xs, ys)
    The graph contains the following edges 1 > 1, 2 > 2

  • merge1(cons(x, xs), cons(y, ys)) → merge1(xs, cons(y, ys))
    The graph contains the following edges 1 > 1, 2 >= 2

  • merge1(cons(x, xs), cons(y, ys)) → merge1(cons(x, xs), ys)
    The graph contains the following edges 1 >= 1, 2 > 2

(50) YES

(51) 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(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(52) 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.

(53) 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(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(54) 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].

list1
list2
list3
hamming

(55) 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(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

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

(56) 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

(57) YES

(58) Obligation:

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

LIST1HAMMING
HAMMINGLIST1
HAMMINGLIST2
LIST2HAMMING
HAMMINGLIST3
LIST3HAMMING

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
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(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hammingapp(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(59) 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.

(60) Obligation:

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

LIST1HAMMING
HAMMINGLIST1
HAMMINGLIST2
LIST2HAMMING
HAMMINGLIST3
LIST3HAMMING

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

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

(61) 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].

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

(62) Obligation:

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

LIST1HAMMING
HAMMINGLIST1
HAMMINGLIST2
LIST2HAMMING
HAMMINGLIST3
LIST3HAMMING

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

(63) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = HAMMING evaluates to t =HAMMING

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]




Rewriting sequence

HAMMINGLIST1
with rule HAMMINGLIST1 at position [] and matcher [ ]

LIST1HAMMING
with rule LIST1HAMMING

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(64) NO