(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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, X)) → MARK(f(a, b))
ACTIVE(f(X, X)) → F(a, b)
ACTIVE(b) → MARK(a)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(a) → ACTIVE(a)
MARK(b) → ACTIVE(b)
F(mark(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
The TRS R consists of the following rules:
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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 2 SCCs with 5 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
The TRS R consists of the following rules:
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- F(X1, mark(X2)) → F(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- F(mark(X1), X2) → F(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- F(active(X1), X2) → F(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- F(X1, active(X2)) → F(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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:
MARK(f(X1, X2)) → MARK(X1)
Used ordering: Polynomial interpretation [POLO]:
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(a) = 0
POL(active(x1)) = x1
POL(b) = 0
POL(f(x1, x2)) = 1 + x1 + x2
POL(mark(x1)) = x1
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(f(a, b))
The TRS R consists of the following rules:
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
MARK(
f(
X1,
X2)) →
ACTIVE(
f(
mark(
X1),
X2)) we obtained the following new rules [LPAR04]:
MARK(f(a, b)) → ACTIVE(f(mark(a), b))
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(mark(a), b))
The TRS R consists of the following rules:
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(mark(a), b))
The TRS R consists of the following rules:
mark(a) → active(a)
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
MARK(
f(
a,
b)) →
ACTIVE(
f(
mark(
a),
b)) at position [0] we obtained the following new rules [LPAR04]:
MARK(f(a, b)) → ACTIVE(f(a, b))
MARK(f(a, b)) → ACTIVE(f(active(a), b))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(a, b))
MARK(f(a, b)) → ACTIVE(f(active(a), b))
The TRS R consists of the following rules:
mark(a) → active(a)
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(a, b)) → ACTIVE(f(active(a), b))
ACTIVE(f(X, X)) → MARK(f(a, b))
The TRS R consists of the following rules:
mark(a) → active(a)
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(a, b)) → ACTIVE(f(active(a), b))
ACTIVE(f(X, X)) → MARK(f(a, b))
The TRS R consists of the following rules:
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
MARK(
f(
a,
b)) →
ACTIVE(
f(
active(
a),
b)) at position [0] we obtained the following new rules [LPAR04]:
MARK(f(a, b)) → ACTIVE(f(a, b))
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(a, b))
The TRS R consists of the following rules:
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(26) TRUE