(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 1
POL(d) = 0
POL(double(x1)) = x1
POL(f(x1, x2)) = x1 + x2
POL(g(x1, x2)) = x1 + x2
POL(h) = 0
POL(half(x1)) = x1
POL(id(x1)) = 2 + x1
POL(s(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(s(0), y) → y
id(x) → f(x, s(0))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
Q is empty.
(3) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The TRS R 2 is
f(s(x), y) → f(half(s(x)), double(y))
The signature Sigma is {
f}
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
G(h, s(s(x))) → G(h, x)
DOUBLE(x) → G(d, x)
HALF(x) → G(h, x)
F(s(x), y) → F(half(s(x)), double(y))
F(s(x), y) → HALF(s(x))
F(s(x), y) → DOUBLE(y)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(h, s(s(x))) → G(h, x)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(h, s(s(x))) → G(h, x)
R is empty.
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(12) 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].
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(h, s(s(x))) → G(h, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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:
- G(h, s(s(x))) → G(h, x)
The graph contains the following edges 1 >= 1, 2 > 2
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
R is empty.
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(19) 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].
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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:
- G(d, s(x)) → G(d, x)
The graph contains the following edges 1 >= 1, 2 > 2
(22) TRUE
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(half(s(x)), double(y))
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(24) 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.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(half(s(x)), double(y))
The TRS R consists of the following rules:
half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(26) 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].
f(s(x0), x1)
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(half(s(x)), double(y))
The TRS R consists of the following rules:
half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
We have to consider all minimal (P,Q,R)-chains.
(28) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(s(x), y) → F(half(s(x)), double(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(F(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(g(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
g(x, 0) → 0
g(h, s(0)) → 0
half(x) → g(h, x)
g(h, s(s(x))) → s(g(h, x))
(29) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
We have to consider all minimal (P,Q,R)-chains.
(30) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(31) TRUE