(0) Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
The relative TRS consists of the following S rules:
rand(x) → rand(s(x))
rand(x) → x
(1) RelTRStoRelADPProof (EQUIVALENT transformation)
We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].
(2) Obligation:
Relative ADP Problem with
absolute ADPs:
f(x, c(y)) → F(x, s(f(y, y)))
f(x, c(y)) → f(x, s(F(y, y)))
f(s(x), y) → F(x, s(c(y)))
and relative ADPs:
rand(x) → RAND(s(x))
rand(x) → x
(3) RelADPDepGraphProof (EQUIVALENT transformation)
We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
2 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 2 subproblems.
(4) Complex Obligation (AND)
(5) Obligation:
Relative ADP Problem with
absolute ADPs:
f(s(x), y) → F(x, s(c(y)))
f(x, c(y)) → f(x, s(f(y, y)))
and relative ADPs:
rand(x) → rand(s(x))
rand(x) → x
(6) RelADPCleverAfsProof (SOUND transformation)
We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.
Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 =
c_1 = 0
f_2 = 1
F_2 = 1
rand_1 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
F(x1, x2) = F(x1)
s(x1) = s(x1)
c(x1) = c
f(x1, x2) = f(x1)
Recursive path order with status [RPO].
Quasi-Precedence:
[F1, s1, c] > f1
Status:
F1: multiset
s1: multiset
c: multiset
f1: [1]
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s0(x)) → F(x)
The TRS R consists of the following rules:
f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)
rand0(x) → x
Q is empty.
We have to consider all (P,Q,R)-chains.
(8) 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:
rand0(x) → x
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1)) = x1
POL(f(x1)) = x1
POL(rand0(x1)) = 1 + x1
POL(s0(x1)) = x1
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s0(x)) → F(x)
The TRS R consists of the following rules:
f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)
Q is empty.
We have to consider all (P,Q,R)-chains.
(10) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(s0(x)) → F(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1) =
x1
s0(
x1) =
s0(
x1)
f(
x1) =
f
rand0(
x1) =
rand0
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s0_1=1
f=1
rand0=2
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)
(11) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x) → f(x)
rand0(x) → rand0(s0(x))
f(s0(x)) → f(x)
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(13) YES
(14) Obligation:
Relative ADP Problem with
absolute ADPs:
f(x, c(y)) → f(x, s(F(y, y)))
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
and relative ADPs:
rand(x) → rand(s(x))
rand(x) → x
(15) RelADPCleverAfsProof (SOUND transformation)
We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.
Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = 0
c_1 =
f_2 = 0
F_2 =
rand_1 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
F(x1, x2) = F(x1, x2)
c(x1) = c(x1)
f(x1, x2) = x2
s(x1) = s
Recursive path order with status [RPO].
Quasi-Precedence:
F2 > s
c1 > s
Status:
F2: [2,1]
c1: multiset
s: multiset
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F0(x, c0(y)) → F0(y, y)
The TRS R consists of the following rules:
f(c0(y)) → f(s)
rand0(x) → rand0(s)
f(y) → f(s)
rand0(x) → x
Q is empty.
We have to consider all (P,Q,R)-chains.
(17) 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:
rand0(x) → x
Used ordering: Polynomial interpretation [POLO]:
POL(F0(x1, x2)) = x1 + 2·x2
POL(c0(x1)) = 2·x1
POL(f(x1)) = x1
POL(rand0(x1)) = 2 + x1
POL(s) = 0
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F0(x, c0(y)) → F0(y, y)
The TRS R consists of the following rules:
f(c0(y)) → f(s)
rand0(x) → rand0(s)
f(y) → f(s)
Q is empty.
We have to consider all (P,Q,R)-chains.
(19) 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:
F0(x, c0(y)) → F0(y, y)
Strictly oriented rules of the TRS R:
f(c0(y)) → f(s)
Used ordering: Polynomial interpretation [POLO]:
POL(F0(x1, x2)) = x1 + 2·x2
POL(c0(x1)) = 2 + 2·x1
POL(f(x1)) = x1
POL(rand0(x1)) = x1
POL(s) = 0
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rand0(x) → rand0(s)
f(y) → f(s)
Q is empty.
We have to consider all (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES