(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
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:
H(z, e(x)) → H(c(z), d(z, x))
H(z, e(x)) → D(z, x)
D(z, g(x, y)) → G(e(x), d(z, y))
D(z, g(x, y)) → D(z, y)
D(c(z), g(g(x, y), 0)) → G(d(c(z), g(x, y)), d(z, g(x, y)))
D(c(z), g(g(x, y), 0)) → D(c(z), g(x, y))
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
G(e(x), e(y)) → G(x, y)
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
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 3 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(e(x), e(y)) → G(x, y)
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
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:
G(e(x), e(y)) → G(x, y)
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:
- G(e(x), e(y)) → G(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(c(z), g(g(x, y), 0)) → D(c(z), g(x, y))
D(z, g(x, y)) → D(z, y)
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
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:
D(c(z), g(g(x, y), 0)) → D(c(z), g(x, y))
D(z, g(x, y)) → D(z, y)
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
The TRS R consists of the following rules:
g(e(x), e(y)) → e(g(x, y))
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:
- D(c(z), g(g(x, y), 0)) → D(c(z), g(x, y))
The graph contains the following edges 1 >= 1, 2 > 2
- D(z, g(x, y)) → D(z, y)
The graph contains the following edges 1 >= 1, 2 > 2
- D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
The graph contains the following edges 1 > 1, 2 > 2
(14) TRUE
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(z, e(x)) → H(c(z), d(z, x))
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
H(
z,
e(
x)) →
H(
c(
z),
d(
z,
x)) we obtained the following new rules [LPAR04]:
H(c(y_0), e(x1)) → H(c(c(y_0)), d(c(y_0), x1))
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(c(y_0), e(x1)) → H(c(c(y_0)), d(c(y_0), x1))
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
H(
c(
y_0),
e(
x1)) →
H(
c(
c(
y_0)),
d(
c(
y_0),
x1)) we obtained the following new rules [LPAR04]:
H(c(c(y_0)), e(x1)) → H(c(c(c(y_0))), d(c(c(y_0)), x1))
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(c(c(y_0)), e(x1)) → H(c(c(c(y_0))), d(c(c(y_0)), x1))
The TRS R consists of the following rules:
h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) → e(g(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.