(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(active(c)) → TOP(mark(c))
TOP(mark(x)) → TOP(check(x))
TOP(mark(x)) → CHECK(x)
CHECK(f(x)) → F(check(x))
CHECK(f(x)) → CHECK(x)
CHECK(x) → START(match(f(X), x))
CHECK(x) → MATCH(f(X), x)
CHECK(x) → F(X)
MATCH(f(x), f(y)) → F(match(x, y))
MATCH(f(x), f(y)) → MATCH(x, y)
MATCH(X, x) → PROPER(x)
PROPER(f(x)) → F(proper(x))
PROPER(f(x)) → PROPER(x)
F(ok(x)) → F(x)
F(found(x)) → F(x)
TOP(found(x)) → TOP(active(x))
TOP(found(x)) → ACTIVE(x)
ACTIVE(f(x)) → F(active(x))
ACTIVE(f(x)) → ACTIVE(x)
F(mark(x)) → F(x)
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 10 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(found(x)) → F(x)
F(ok(x)) → F(x)
F(mark(x)) → F(x)
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(found(x)) → F(x)
F(ok(x)) → F(x)
F(mark(x)) → F(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- F(found(x)) → F(x)
The graph contains the following edges 1 > 1
- F(ok(x)) → F(x)
The graph contains the following edges 1 > 1
- F(mark(x)) → F(x)
The graph contains the following edges 1 > 1
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(x)) → ACTIVE(x)
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(x)) → ACTIVE(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- ACTIVE(f(x)) → ACTIVE(x)
The graph contains the following edges 1 > 1
(14) TRUE
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PROPER(f(x)) → PROPER(x)
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) 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.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PROPER(f(x)) → PROPER(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- PROPER(f(x)) → PROPER(x)
The graph contains the following edges 1 > 1
(19) TRUE
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MATCH(f(x), f(y)) → MATCH(x, y)
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(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:
MATCH(f(x), f(y)) → MATCH(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- MATCH(f(x), f(y)) → MATCH(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(24) TRUE
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHECK(f(x)) → CHECK(x)
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) 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.
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHECK(f(x)) → CHECK(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- CHECK(f(x)) → CHECK(x)
The graph contains the following edges 1 > 1
(29) TRUE
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
TOP(active(c)) → TOP(mark(c))
The TRS R consists of the following rules:
active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(31) 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.
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
TOP(active(c)) → TOP(mark(c))
The TRS R consists of the following rules:
active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
start(ok(x)) → found(x)
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
TOP(active(c)) → TOP(mark(c))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(match(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
f(ok(x)) → ok(f(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
f(found(x)) → found(f(x))
start(ok(x)) → found(x)
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
active(f(x)) → mark(x)
active(f(x)) → f(active(x))
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
The TRS R consists of the following rules:
active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
start(ok(x)) → found(x)
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(35) 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:
match(f(x), f(y)) → f(match(x, y))
Used ordering: Polynomial interpretation [POLO]:
POL(TOP(x1)) = x1
POL(X) = 0
POL(active(x1)) = x1
POL(c) = 0
POL(check(x1)) = 1 + x1
POL(f(x1)) = 1 + x1
POL(found(x1)) = x1
POL(mark(x1)) = 1 + x1
POL(match(x1, x2)) = x1 + x2
POL(ok(x1)) = x1
POL(proper(x1)) = x1
POL(start(x1)) = x1
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
The TRS R consists of the following rules:
active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
start(ok(x)) → found(x)
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(37) 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.
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
The TRS R consists of the following rules:
active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(39) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
Strictly oriented rules of the TRS R:
active(f(x)) → mark(x)
check(x) → start(match(f(X), x))
Used ordering: Polynomial interpretation [POLO]:
POL(TOP(x1)) = x1
POL(X) = 0
POL(active(x1)) = 3 + x1
POL(check(x1)) = 1 + x1
POL(f(x1)) = x1
POL(found(x1)) = 4 + x1
POL(mark(x1)) = 2 + x1
POL(match(x1, x2)) = x1 + x2
POL(ok(x1)) = x1
POL(start(x1)) = x1
(40) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(41) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(42) TRUE