(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
-1(s(x), s(y)) → -1(x, y)
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
GCD(s(x), s(y)) → -1(s(max(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
R is empty.
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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:
- -1(s(x), s(y)) → -1(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
R is empty.
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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:
- MAX(s(x), s(y)) → MAX(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(20) TRUE
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(22) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
R is empty.
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(24) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) 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:
- MIN(s(x), s(y)) → MIN(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(27) TRUE
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(29) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
We have to consider all minimal (P,Q,R)-chains.
(31) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(33) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
GCD(
s(
x),
s(
y)) →
GCD(
-(
s(
max(
x,
y)),
s(
min(
x,
y))),
s(
min(
x,
y))) at position [0] we obtained the following new rules [LPAR04]:
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(35) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(GCD(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(-(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(max(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(min(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
-(x, 0) → x
min(0, y) → 0
min(x, 0) → 0
-(s(x), s(y)) → -(x, y)
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
max(0, y) → y
(36) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(s(x), s(y)) → -(x, y)
-(x, 0) → x
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(37) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(38) TRUE