(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxRelTRS could be proven to be
BOUNDS(1, n^2).
The TRS R consists of the following rules:
ordered(Cons(x', Cons(x, xs))) → ordered[Ite](<(x', x), Cons(x', Cons(x, xs)))
ordered(Cons(x, Nil)) → True
ordered(Nil) → True
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
goal(xs) → ordered(xs)
The (relative) TRS S consists of the following rules:
<(S(x), S(y)) → <(x, y)
<(0, S(y)) → True
<(x, 0) → False
ordered[Ite](True, Cons(x', Cons(x, xs))) → ordered(xs)
ordered[Ite](False, xs) → False
Rewrite Strategy: INNERMOST
(1) RelTrsToTrsProof (UPPER BOUND(ID) transformation)
transformed relative TRS to TRS
(2) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^2).
The TRS R consists of the following rules:
ordered(Cons(x', Cons(x, xs))) → ordered[Ite](<(x', x), Cons(x', Cons(x, xs)))
ordered(Cons(x, Nil)) → True
ordered(Nil) → True
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
goal(xs) → ordered(xs)
<(S(x), S(y)) → <(x, y)
<(0, S(y)) → True
<(x, 0) → False
ordered[Ite](True, Cons(x', Cons(x, xs))) → ordered(xs)
ordered[Ite](False, xs) → False
Rewrite Strategy: INNERMOST
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
ordered(Cons(z0, Cons(z1, z2))) → ordered[Ite](<(z0, z1), Cons(z0, Cons(z1, z2)))
ordered(Cons(z0, Nil)) → True
ordered(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0) → ordered(z0)
<(S(z0), S(z1)) → <(z0, z1)
<(0, S(z0)) → True
<(z0, 0) → False
ordered[Ite](True, Cons(z0, Cons(z1, z2))) → ordered(z2)
ordered[Ite](False, z0) → False
Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
ORDERED(Cons(z0, Nil)) → c1
ORDERED(Nil) → c2
NOTEMPTY(Cons(z0, z1)) → c3
NOTEMPTY(Nil) → c4
GOAL(z0) → c5(ORDERED(z0))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
<'(0, S(z0)) → c7
<'(z0, 0) → c8
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
ORDERED[ITE](False, z0) → c10
S tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
ORDERED(Cons(z0, Nil)) → c1
ORDERED(Nil) → c2
NOTEMPTY(Cons(z0, z1)) → c3
NOTEMPTY(Nil) → c4
GOAL(z0) → c5(ORDERED(z0))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
<'(0, S(z0)) → c7
<'(z0, 0) → c8
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
ORDERED[ITE](False, z0) → c10
K tuples:none
Defined Rule Symbols:
ordered, notEmpty, goal, <, ordered[Ite]
Defined Pair Symbols:
ORDERED, NOTEMPTY, GOAL, <', ORDERED[ITE]
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
GOAL(z0) → c5(ORDERED(z0))
Removed 7 trailing nodes:
NOTEMPTY(Cons(z0, z1)) → c3
<'(0, S(z0)) → c7
NOTEMPTY(Nil) → c4
ORDERED(Nil) → c2
ORDERED(Cons(z0, Nil)) → c1
ORDERED[ITE](False, z0) → c10
<'(z0, 0) → c8
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
ordered(Cons(z0, Cons(z1, z2))) → ordered[Ite](<(z0, z1), Cons(z0, Cons(z1, z2)))
ordered(Cons(z0, Nil)) → True
ordered(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0) → ordered(z0)
<(S(z0), S(z1)) → <(z0, z1)
<(0, S(z0)) → True
<(z0, 0) → False
ordered[Ite](True, Cons(z0, Cons(z1, z2))) → ordered(z2)
ordered[Ite](False, z0) → False
Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
S tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
K tuples:none
Defined Rule Symbols:
ordered, notEmpty, goal, <, ordered[Ite]
Defined Pair Symbols:
ORDERED, <', ORDERED[ITE]
Compound Symbols:
c, c6, c9
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
ordered(Cons(z0, Cons(z1, z2))) → ordered[Ite](<(z0, z1), Cons(z0, Cons(z1, z2)))
ordered(Cons(z0, Nil)) → True
ordered(Nil) → True
notEmpty(Cons(z0, z1)) → True
notEmpty(Nil) → False
goal(z0) → ordered(z0)
ordered[Ite](True, Cons(z0, Cons(z1, z2))) → ordered(z2)
ordered[Ite](False, z0) → False
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
<(S(z0), S(z1)) → <(z0, z1)
<(0, S(z0)) → True
<(z0, 0) → False
Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
S tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
K tuples:none
Defined Rule Symbols:
<
Defined Pair Symbols:
ORDERED, <', ORDERED[ITE]
Compound Symbols:
c, c6, c9
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
We considered the (Usable) Rules:none
And the Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [3]
POL(<(x1, x2)) = [3] + [2]x1 + [3]x2
POL(<'(x1, x2)) = 0
POL(Cons(x1, x2)) = [1] + x2
POL(False) = [1]
POL(ORDERED(x1)) = [1] + x1
POL(ORDERED[ITE](x1, x2)) = x2
POL(S(x1)) = [3] + x1
POL(True) = [1]
POL(c(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c9(x1)) = x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
<(S(z0), S(z1)) → <(z0, z1)
<(0, S(z0)) → True
<(z0, 0) → False
Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
S tuples:
<'(S(z0), S(z1)) → c6(<'(z0, z1))
K tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
Defined Rule Symbols:
<
Defined Pair Symbols:
ORDERED, <', ORDERED[ITE]
Compound Symbols:
c, c6, c9
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
<'(S(z0), S(z1)) → c6(<'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(<(x1, x2)) = 0
POL(<'(x1, x2)) = x2
POL(Cons(x1, x2)) = [2] + x1 + x2
POL(False) = [2]
POL(ORDERED(x1)) = [1] + x1 + [2]x12
POL(ORDERED[ITE](x1, x2)) = [2]x22
POL(S(x1)) = [1] + x1
POL(True) = 0
POL(c(x1, x2)) = x1 + x2
POL(c6(x1)) = x1
POL(c9(x1)) = x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
<(S(z0), S(z1)) → <(z0, z1)
<(0, S(z0)) → True
<(z0, 0) → False
Tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
S tuples:none
K tuples:
ORDERED(Cons(z0, Cons(z1, z2))) → c(ORDERED[ITE](<(z0, z1), Cons(z0, Cons(z1, z2))), <'(z0, z1))
ORDERED[ITE](True, Cons(z0, Cons(z1, z2))) → c9(ORDERED(z2))
<'(S(z0), S(z1)) → c6(<'(z0, z1))
Defined Rule Symbols:
<
Defined Pair Symbols:
ORDERED, <', ORDERED[ITE]
Compound Symbols:
c, c6, c9
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)