(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
serve'(old(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)
Q is empty.
(3) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
serve'(old(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(check(x1)) = x1
POL(free(x1)) = x1
POL(new(x1)) = x1
POL(old(x1)) = 1 + x1
POL(serve) = 0
POL(top(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
old(serve) → free(serve)
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(free(x)) → TOP(check(new(x)))
TOP(free(x)) → CHECK(new(x))
TOP(free(x)) → NEW(x)
NEW(free(x)) → NEW(x)
OLD(free(x)) → OLD(x)
CHECK(free(x)) → CHECK(x)
CHECK(new(x)) → NEW(check(x))
CHECK(new(x)) → CHECK(x)
CHECK(old(x)) → OLD(check(x))
CHECK(old(x)) → CHECK(x)
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.
(10) Complex Obligation (AND)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
OLD(free(x)) → OLD(x)
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
OLD(free(x)) → OLD(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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:
OLD(free(x)) → OLD(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(OLD(x1)) = 2·x1
POL(free(x1)) = 2·x1
(15) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) TRUE
(18) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
OLD(free(x)) → OLD(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
NEW(free(x)) → NEW(x)
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
NEW(free(x)) → NEW(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
NEW(free(x)) → NEW(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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:
NEW(free(x)) → NEW(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(NEW(x1)) = 2·x1
POL(free(x1)) = 2·x1
(26) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(28) TRUE
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHECK(new(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHECK(new(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(32) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHECK(new(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(34) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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:
CHECK(new(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(CHECK(x1)) = 2·x1
POL(free(x1)) = 2·x1
POL(new(x1)) = 2·x1
POL(old(x1)) = 2·x1
(35) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(36) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(37) TRUE
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(free(x)) → TOP(check(new(x)))
The TRS R consists of the following rules:
top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(39) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(free(x)) → TOP(check(new(x)))
The TRS R consists of the following rules:
new(free(x)) → free(new(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
old(free(x)) → free(old(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(41) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
old(free(x)) → free(old(x))
Used ordering: Polynomial interpretation [POLO]:
POL(TOP(x1)) = x1
POL(check(x1)) = x1
POL(free(x1)) = 1 + x1
POL(new(x1)) = 1 + x1
POL(old(x1)) = 2 + 2·x1
POL(serve) = 0
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(free(x)) → TOP(check(new(x)))
The TRS R consists of the following rules:
new(free(x)) → free(new(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(43) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
check(old(x)) → old(check(x))
check(old(x)) → old(x)
Used ordering: Polynomial interpretation [POLO]:
POL(TOP(x1)) = 2·x1
POL(check(x1)) = 2·x1
POL(free(x1)) = 2·x1
POL(new(x1)) = x1
POL(old(x1)) = 2 + 2·x1
POL(serve) = 0
(44) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(free(x)) → TOP(check(new(x)))
The TRS R consists of the following rules:
new(free(x)) → free(new(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(45) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
TOP(free(x)) → TOP(check(new(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
The following usable rules [FROCOS05] were oriented:
new(free(x)) → free(new(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
(46) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
new(free(x)) → free(new(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(47) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(48) TRUE
(49) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(50) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(free(x)) → TOP(check(new(x)))
The TRS R consists of the following rules:
new(free(x)) → free(new(x))
new(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)
old(free(x)) → free(old(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.