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:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
The TRS R 2 is
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The signature Sigma is {true, false, less_leaves}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
SHUFFLE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(w, z)
CONCAT(cons(u, v), y) → CONCAT(v, y)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
QUOT(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
SHUFFLE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(w, z)
CONCAT(cons(u, v), y) → CONCAT(v, y)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
QUOT(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 7 SCCs with 5 less nodes.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONCAT(cons(u, v), y) → CONCAT(v, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONCAT(cons(u, v), y) → CONCAT(v, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONCAT(cons(u, v), y) → CONCAT(v, y)
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:
- CONCAT(cons(u, v), y) → CONCAT(v, y)
The graph contains the following edges 1 > 1, 2 >= 2
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
The TRS R consists of the following rules:
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
The TRS R consists of the following rules:
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
The set Q consists of the following terms:
concat(leaf, x0)
concat(cons(x0, x1), x2)
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.
The following rules are removed from R:
concat(leaf, y) → y
Used ordering: POLO with Polynomial interpretation [25]:
POL(LESS_LEAVES(x1, x2)) = 2·x1 + 2·x2
POL(concat(x1, x2)) = 2 + x1 + x2
POL(cons(x1, x2)) = 2 + 2·x1 + x2
POL(leaf) = 0
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
The TRS R consists of the following rules:
concat(cons(u, v), y) → cons(u, concat(v, y))
The set Q consists of the following terms:
concat(leaf, x0)
concat(cons(x0, x1), x2)
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:
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
Strictly oriented rules of the TRS R:
concat(cons(u, v), y) → cons(u, concat(v, y))
Used ordering: POLO with Polynomial interpretation [25]:
POL(LESS_LEAVES(x1, x2)) = 2·x1 + x2
POL(concat(x1, x2)) = 1 + 2·x1 + x2
POL(cons(x1, x2)) = 2 + 2·x1 + x2
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
concat(leaf, x0)
concat(cons(x0, x1), x2)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP(add(n, x), y) → APP(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP(add(n, x), y) → APP(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP(add(n, x), y) → APP(x, y)
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:
- APP(add(n, x), y) → APP(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REVERSE(add(n, x)) → REVERSE(x)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REVERSE(add(n, x)) → REVERSE(x)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REVERSE(add(n, x)) → REVERSE(x)
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:
- REVERSE(add(n, x)) → REVERSE(x)
The graph contains the following edges 1 > 1
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
The set Q consists of the following terms:
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
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:
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
Used ordering: POLO with Polynomial interpretation [25]:
POL(SHUFFLE(x1)) = x1
POL(add(x1, x2)) = 1 + 2·x1 + x2
POL(app(x1, x2)) = x1 + x2
POL(nil) = 0
POL(reverse(x1)) = x1
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
The set Q consists of the following terms:
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
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:
- MINUS(s(x), s(y)) → MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
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.
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( minus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( QUOT(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
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:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.