(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(x)) → F(f(x))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(X))))))
CHK(no(f(x))) → F(f(f(f(f(X)))))
CHK(no(f(x))) → F(f(f(f(X))))
CHK(no(f(x))) → F(f(f(X)))
CHK(no(f(x))) → F(f(X))
CHK(no(f(x))) → F(X)
MAT(f(x), f(y)) → F(mat(x, y))
MAT(f(x), f(y)) → MAT(x, y)
CHK(no(c)) → ACTIVE(c)
F(active(x)) → ACTIVE(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)
TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(X))))))))
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
TP(mark(x)) → F(f(f(f(f(f(X))))))
TP(mark(x)) → F(f(f(f(f(X)))))
TP(mark(x)) → F(f(f(f(X))))
TP(mark(x)) → F(f(f(X)))
TP(mark(x)) → F(f(X))
TP(mark(x)) → F(X)
The TRS R consists of the following rules:
active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 27 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)
The TRS R consists of the following rules:
active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)
The TRS R consists of the following rules:
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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 dependency pairs:
F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)
Strictly oriented rules of the TRS R:
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))
Used ordering: Knuth-Bendix order [KBO] with precedence:
ACTIVE1 > f1 > active1 > mark1 > F1 > no1
and weight map:
f_1=2
active_1=3
no_1=1
mark_1=1
F_1=1
ACTIVE_1=1
The variable weight is 1
(9) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
The TRS R consists of the following rules:
active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
The TRS R consists of the following rules:
mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
The TRS R consists of the following rules:
mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
The set Q consists of the following terms:
mat(f(x0), f(y))
mat(f(x0), c)
We have to consider all minimal (P,Q,R)-chains.
(17) 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 dependency pairs:
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
Strictly oriented rules of the TRS R:
mat(f(x), c) → no(c)
Used ordering: Polynomial interpretation [POLO]:
POL(CHK(x1)) = 2·x1
POL(X) = 0
POL(c) = 2
POL(f(x1)) = 2·x1
POL(mat(x1, x2)) = x1 + 2·x2
POL(no(x1)) = 1 + x1
POL(y) = 0
(18) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
mat(f(x), f(y)) → f(mat(x, y))
The set Q consists of the following terms:
mat(f(x0), f(y))
mat(f(x0), c)
We have to consider all minimal (P,Q,R)-chains.
(19) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(20) YES
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
The TRS R consists of the following rules:
active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(22) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
The TRS R consists of the following rules:
mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) → active(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(24) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
TP(
mark(
x)) →
TP(
chk(
mat(
f(
f(
f(
f(
f(
f(
f(
f(
f(
f(
X)))))))))),
x))) at position [0] we obtained the following new rules [LPAR04]:
TP(mark(f(y))) → TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))) → TP(mark(f(y))) → TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
TP(mark(c)) → TP(chk(no(c))) → TP(mark(c)) → TP(chk(no(c)))
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TP(mark(f(y))) → TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
TP(mark(c)) → TP(chk(no(c)))
The TRS R consists of the following rules:
mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) → active(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(27) TRUE