Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
PROPER(h(X)) → H(proper(X))
ACTIVE(g(X)) → H(X)
PROPER(g(X)) → PROPER(X)
H(ok(X)) → H(X)
TOP(mark(X)) → PROPER(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
PROPER(g(X)) → G(proper(X))
ACTIVE(h(d)) → G(c)
TOP(ok(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
PROPER(h(X)) → PROPER(X)
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
PROPER(h(X)) → H(proper(X))
ACTIVE(g(X)) → H(X)
PROPER(g(X)) → PROPER(X)
H(ok(X)) → H(X)
TOP(mark(X)) → PROPER(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
PROPER(g(X)) → G(proper(X))
ACTIVE(h(d)) → G(c)
TOP(ok(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
PROPER(h(X)) → PROPER(X)
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(ok(X)) → H(X)
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(ok(X)) → H(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
H(ok(X)) → H(X)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(H(x1)) = 2·x1
POL(ok(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
H(ok(X)) → H(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(ok(X)) → G(X)
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(ok(X)) → G(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
G(ok(X)) → G(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
G(ok(X)) → G(X)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(G(x1)) = 2·x1
POL(ok(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(PROPER(x1)) = 2·x1
POL(g(x1)) = 2·x1
POL(h(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(mark(X)) → TOP(proper(X)) at position [0] we obtained the following new rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(c)) → TOP(ok(c))
TOP(mark(d)) → TOP(ok(d))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(mark(c)) → TOP(ok(c))
TOP(ok(X)) → TOP(active(X))
TOP(mark(d)) → TOP(ok(d))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(ok(X)) → TOP(active(X))
TOP(mark(d)) → TOP(ok(d))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:
TOP(ok(c)) → TOP(mark(d))
TOP(ok(g(x0))) → TOP(mark(h(x0)))
TOP(ok(h(d))) → TOP(mark(g(c)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(c)) → TOP(mark(d))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(ok(g(x0))) → TOP(mark(h(x0)))
TOP(ok(h(d))) → TOP(mark(g(c)))
TOP(mark(d)) → TOP(ok(d))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(ok(g(x0))) → TOP(mark(h(x0)))
TOP(ok(h(d))) → TOP(mark(g(c)))
The TRS R consists of the following rules:
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(ok(g(x0))) → TOP(mark(h(x0)))
TOP(ok(h(d))) → TOP(mark(g(c)))
The TRS R consists of the following rules:
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
TOP(ok(h(d))) → TOP(mark(g(c)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(TOP(x1)) = 2·x1
POL(c) = 0
POL(d) = 2
POL(g(x1)) = 2·x1
POL(h(x1)) = 2·x1
POL(mark(x1)) = 2·x1
POL(ok(x1)) = 2·x1
POL(proper(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
TOP(ok(g(x0))) → TOP(mark(h(x0)))
The TRS R consists of the following rules:
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
TOP(ok(g(x0))) → TOP(mark(h(x0)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(TOP(x1)) = 2·x1
POL(c) = 2
POL(d) = 0
POL(g(x1)) = 2 + 2·x1
POL(h(x1)) = x1
POL(mark(x1)) = 2 + 2·x1
POL(ok(x1)) = 2 + 2·x1
POL(proper(x1)) = 2 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(h(x0))) → TOP(h(proper(x0)))
The TRS R consists of the following rules:
h(ok(X)) → ok(h(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.
We have reversed the following QTRS:
The set of rules R is
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
The set Q is empty.
We have obtained the following QTRS:
g(active(x)) → h(mark(x))
c'(active(x)) → d'(mark(x))
d'(h(active(x))) → c'(g(mark(x)))
g(proper(x)) → proper(g(x))
h(proper(x)) → proper(h(x))
c'(proper(x)) → c'(ok(x))
d'(proper(x)) → d'(ok(x))
ok(g(x)) → g(ok(x))
ok(h(x)) → h(ok(x))
mark(top(x)) → proper(top(x))
ok(top(x)) → active(top(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
g(active(x)) → h(mark(x))
c'(active(x)) → d'(mark(x))
d'(h(active(x))) → c'(g(mark(x)))
g(proper(x)) → proper(g(x))
h(proper(x)) → proper(h(x))
c'(proper(x)) → c'(ok(x))
d'(proper(x)) → d'(ok(x))
ok(g(x)) → g(ok(x))
ok(h(x)) → h(ok(x))
mark(top(x)) → proper(top(x))
ok(top(x)) → active(top(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
The set Q is empty.
We have obtained the following QTRS:
g(active(x)) → h(mark(x))
c'(active(x)) → d'(mark(x))
d'(h(active(x))) → c'(g(mark(x)))
g(proper(x)) → proper(g(x))
h(proper(x)) → proper(h(x))
c'(proper(x)) → c'(ok(x))
d'(proper(x)) → d'(ok(x))
ok(g(x)) → g(ok(x))
ok(h(x)) → h(ok(x))
mark(top(x)) → proper(top(x))
ok(top(x)) → active(top(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
g(active(x)) → h(mark(x))
c'(active(x)) → d'(mark(x))
d'(h(active(x))) → c'(g(mark(x)))
g(proper(x)) → proper(g(x))
h(proper(x)) → proper(h(x))
c'(proper(x)) → c'(ok(x))
d'(proper(x)) → d'(ok(x))
ok(g(x)) → g(ok(x))
ok(h(x)) → h(ok(x))
mark(top(x)) → proper(top(x))
ok(top(x)) → active(top(x))
Q is empty.