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(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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:
ACTIVE(f(b, X, c)) → F(X, c, X)
ACTIVE(f(X1, X2, X3)) → F(X1, active(X2), X3)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
TOP(mark(X)) → PROPER(X)
F(X1, mark(X2), X3) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(b, X, c)) → F(X, c, X)
ACTIVE(f(X1, X2, X3)) → F(X1, active(X2), X3)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
TOP(mark(X)) → PROPER(X)
F(X1, mark(X2), X3) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
The TRS R consists of the following rules:
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 > 2, 3 > 3
- F(X1, mark(X2), X3) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
The TRS R consists of the following rules:
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- PROPER(f(X1, X2, X3)) → PROPER(X1)
The graph contains the following edges 1 > 1
- PROPER(f(X1, X2, X3)) → PROPER(X2)
The graph contains the following edges 1 > 1
- PROPER(f(X1, X2, X3)) → PROPER(X3)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)
The TRS R consists of the following rules:
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- ACTIVE(f(X1, X2, X3)) → ACTIVE(X2)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
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(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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.
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.
No dependency pairs are removed.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(TOP(x1)) = 2·x1
POL(active(x1)) = 2·x1
POL(b) = 0
POL(c) = 0
POL(f(x1, x2, x3)) = x1 + 2·x2 + 2·x3
POL(mark(x1)) = x1
POL(ok(x1)) = 2·x1
POL(proper(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
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:
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
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(b)) → TOP(ok(b))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(c)) → TOP(ok(c))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(c)) → TOP(ok(c))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
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
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(b)) → TOP(ok(b))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
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(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(ok(f(b, x0, c))) → TOP(mark(f(x0, c, x0)))
TOP(ok(c)) → TOP(mark(b))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(f(b, x0, c))) → TOP(mark(f(x0, c, x0)))
TOP(ok(c)) → TOP(mark(b))
TOP(mark(b)) → TOP(ok(b))
The TRS R consists of the following rules:
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
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
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(f(b, x0, c))) → TOP(mark(f(x0, c, x0)))
The TRS R consists of the following rules:
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
TOP(ok(f(b, x0, c))) → TOP(mark(f(x0, c, x0)))
The remaining pairs can at least be oriented weakly.
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( f(x1, ..., x3) ) = | | + | | · | x1 | + | | · | x2 | + | | · | x3 |
Tuple symbols:
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(c) → ok(c)
proper(b) → ok(b)
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ SemLabProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, active(x1), x2))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
The TRS R consists of the following rules:
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.active: 0
c: 0
f: x1
mark: 0
ok: 0
b: 1
TOP: 0
proper: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
TOP.0(mark.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
The TRS R consists of the following rules:
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
proper.0(c.) → ok.0(c.)
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
TOP.0(mark.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
The TRS R consists of the following rules:
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
proper.0(c.) → ok.0(c.)
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
TOP.0(mark.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.1(x1), proper.0(x2)))
The remaining pairs can at least be oriented weakly.
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
Used ordering: Polynomial interpretation [25]:
POL(TOP.0(x1)) = x1
POL(active.0(x1)) = x1
POL(active.1(x1)) = x1
POL(b.) = 0
POL(c.) = 1
POL(f.0-0-0(x1, x2, x3)) = x2 + x3
POL(f.0-0-1(x1, x2, x3)) = x2 + x3
POL(f.0-1-0(x1, x2, x3)) = x2 + x3
POL(f.0-1-1(x1, x2, x3)) = x2 + x3
POL(f.1-0-0(x1, x2, x3)) = x2 + x3
POL(f.1-0-1(x1, x2, x3)) = x2 + x3
POL(f.1-1-0(x1, x2, x3)) = x2 + x3
POL(f.1-1-1(x1, x2, x3)) = x2 + x3
POL(mark.0(x1)) = x1
POL(mark.1(x1)) = 1 + x1
POL(ok.0(x1)) = x1
POL(ok.1(x1)) = x1
POL(proper.0(x1)) = x1
POL(proper.1(x1)) = x1
The following usable rules [17] were oriented:
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
proper.0(c.) → ok.0(c.)
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
The TRS R consists of the following rules:
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
proper.0(c.) → ok.0(c.)
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
The remaining pairs can at least be oriented weakly.
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
Used ordering: Polynomial interpretation [25]:
POL(TOP.0(x1)) = x1
POL(active.0(x1)) = x1
POL(active.1(x1)) = x1
POL(b.) = 1
POL(c.) = 0
POL(f.0-0-0(x1, x2, x3)) = x1 + x2
POL(f.0-0-1(x1, x2, x3)) = x1 + x2
POL(f.0-1-0(x1, x2, x3)) = x1 + x2
POL(f.0-1-1(x1, x2, x3)) = x1 + x2
POL(f.1-0-0(x1, x2, x3)) = x1 + x2
POL(f.1-0-1(x1, x2, x3)) = x1 + x2
POL(f.1-1-0(x1, x2, x3)) = x1 + x2
POL(f.1-1-1(x1, x2, x3)) = x1 + x2
POL(mark.0(x1)) = 1 + x1
POL(mark.1(x1)) = 0
POL(ok.0(x1)) = x1
POL(ok.1(x1)) = x1
POL(proper.0(x1)) = x1
POL(proper.1(x1)) = x1
The following usable rules [17] were oriented:
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
proper.0(c.) → ok.0(c.)
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
The TRS R consists of the following rules:
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
proper.0(c.) → ok.0(c.)
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.0(x1), x2))
TOP.0(ok.1(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.1(x1), x2))
TOP.0(ok.1(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-0-1(x0, active.1(x1), x2))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.0(x1), x2))
TOP.0(ok.1(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.1(x1), x2))
TOP.0(ok.1(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, active.1(x1), x2))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, active.0(x1), x2))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(x0, active.0(x1), x2))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:
POL(TOP.0(x1)) = x1
POL(active.0(x1)) = 0
POL(active.1(x1)) = 0
POL(b.) = 0
POL(c.) = 0
POL(f.0-0-0(x1, x2, x3)) = 1 + x1 + x3
POL(f.0-0-1(x1, x2, x3)) = 1 + x1 + x3
POL(f.0-1-0(x1, x2, x3)) = 1 + x1 + x3
POL(f.0-1-1(x1, x2, x3)) = 1 + x1 + x3
POL(f.1-0-0(x1, x2, x3)) = 1
POL(f.1-0-1(x1, x2, x3)) = 1
POL(f.1-1-0(x1, x2, x3)) = 1
POL(f.1-1-1(x1, x2, x3)) = 1
POL(mark.0(x1)) = 0
POL(mark.1(x1)) = 0
POL(ok.0(x1)) = 1 + x1
POL(ok.1(x1)) = 1 + x1
The following usable rules [17] were oriented:
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active.1(f.1-1-0(X1, X2, X3)) → f.1-0-0(X1, active.1(X2), X3)
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
proper.0(c.) → ok.0(c.)
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, active.0(X2), X3)
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
active.0(f.1-0-0(b., X, c.)) → mark.0(f.0-0-0(X, c., X))
proper.1(b.) → ok.1(b.)
active.1(f.1-1-1(X1, X2, X3)) → f.1-0-1(X1, active.1(X2), X3)
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(X1, active.0(X2), X3)
active.1(f.1-1-0(b., X, c.)) → mark.0(f.1-0-1(X, c., X))
f.0-0-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.1(f.1-1-0(X1, X2, X3))
f.1-0-0(X1, mark.0(X2), X3) → mark.0(f.1-0-0(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
proper.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.0(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(X1, active.0(X2), X3)
active.0(c.) → mark.1(b.)
f.0-0-0(X1, mark.1(X2), X3) → mark.1(f.0-1-0(X1, X2, X3))
f.0-0-1(X1, mark.0(X2), X3) → mark.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.1(f.0-1-0(X1, X2, X3))
f.1-0-1(X1, mark.0(X2), X3) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.1(f.1-1-1(X1, X2, X3))
f.1-0-0(X1, mark.1(X2), X3) → mark.1(f.1-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, active.0(X2), X3)
active.1(f.0-1-0(X1, X2, X3)) → f.0-0-0(X1, active.1(X2), X3)
proper.1(f.1-1-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.1(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(X1, mark.1(X2), X3) → mark.1(f.1-1-1(X1, X2, X3))
f.0-0-0(X1, mark.0(X2), X3) → mark.0(f.0-0-0(X1, X2, X3))
active.1(f.0-1-1(X1, X2, X3)) → f.0-0-1(X1, active.1(X2), X3)
proper.1(f.1-1-0(X1, X2, X3)) → f.0-0-0(proper.1(X1), proper.1(X2), proper.0(X3))
proper.1(f.0-1-1(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.1(X2), proper.1(X3))
f.0-0-1(X1, mark.1(X2), X3) → mark.1(f.0-1-1(X1, X2, X3))
f.0-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
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.