0.00/0.85
0.00/0.85
(0) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
lte(Cons(x', xs'), Cons(x, xs)) → lte(xs', xs)
0.00/0.85
lte(Cons(x, xs), Nil) → False
0.00/0.85
even(Cons(x, Nil)) → False
0.00/0.85
even(Cons(x', Cons(x, xs))) → even(xs)
0.00/0.85
notEmpty(Cons(x, xs)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
lte(Nil, y) → True
0.00/0.85
even(Nil) → True
0.00/0.85
goal(x, y) → and(lte(x, y), even(x))
The (relative) TRS S consists of the following rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
Rewrite Strategy: INNERMOST
0.00/0.85
0.00/0.85
(1) CpxRelTrsToCDT (UPPER BOUND (ID) transformation)
Relative innermost TRS to CDT Problem.
0.00/0.85
0.00/0.85
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3)
0.00/0.85
lte(Cons(z0, z1), Nil) → False
0.00/0.85
lte(Nil, z0) → True
0.00/0.85
even(Cons(z0, Nil)) → False
0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2)
0.00/0.85
even(Nil) → True
0.00/0.85
notEmpty(Cons(z0, z1)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
0.00/0.85
GOAL(z0, z1) → c12(AND(lte(z0, z1), even(z0)), LTE(z0, z1), EVEN(z0))
S tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
0.00/0.85
GOAL(z0, z1) → c12(AND(lte(z0, z1), even(z0)), LTE(z0, z1), EVEN(z0))
K tuples:none
Defined Rule Symbols:
lte, even, notEmpty, goal, and
Defined Pair Symbols:
LTE, EVEN, GOAL
Compound Symbols:
c4, c8, c12
0.00/0.85
0.00/0.85
(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
0.00/0.85
0.00/0.85
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3)
0.00/0.85
lte(Cons(z0, z1), Nil) → False
0.00/0.85
lte(Nil, z0) → True
0.00/0.85
even(Cons(z0, Nil)) → False
0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2)
0.00/0.85
even(Nil) → True
0.00/0.85
notEmpty(Cons(z0, z1)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
0.00/0.85
GOAL(z0, z1) → c12(LTE(z0, z1), EVEN(z0))
S tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
0.00/0.85
GOAL(z0, z1) → c12(LTE(z0, z1), EVEN(z0))
K tuples:none
Defined Rule Symbols:
lte, even, notEmpty, goal, and
Defined Pair Symbols:
LTE, EVEN, GOAL
Compound Symbols:
c4, c8, c12
0.00/0.85
0.00/0.85
(5) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)
Split RHS of tuples not part of any SCC
0.00/0.85
0.00/0.85
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3)
0.00/0.85
lte(Cons(z0, z1), Nil) → False
0.00/0.85
lte(Nil, z0) → True
0.00/0.85
even(Cons(z0, Nil)) → False
0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2)
0.00/0.85
even(Nil) → True
0.00/0.85
notEmpty(Cons(z0, z1)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
0.00/0.85
GOAL(z0, z1) → c(LTE(z0, z1))
0.00/0.85
GOAL(z0, z1) → c(EVEN(z0))
S tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
0.00/0.85
GOAL(z0, z1) → c(LTE(z0, z1))
0.00/0.85
GOAL(z0, z1) → c(EVEN(z0))
K tuples:none
Defined Rule Symbols:
lte, even, notEmpty, goal, and
Defined Pair Symbols:
LTE, EVEN, GOAL
Compound Symbols:
c4, c8, c
0.00/0.85
0.00/0.85
(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 2 leading nodes:
GOAL(z0, z1) → c(LTE(z0, z1))
0.00/0.85
GOAL(z0, z1) → c(EVEN(z0))
0.00/0.85
0.00/0.85
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3)
0.00/0.85
lte(Cons(z0, z1), Nil) → False
0.00/0.85
lte(Nil, z0) → True
0.00/0.85
even(Cons(z0, Nil)) → False
0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2)
0.00/0.85
even(Nil) → True
0.00/0.85
notEmpty(Cons(z0, z1)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
S tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
K tuples:none
Defined Rule Symbols:
lte, even, notEmpty, goal, and
Defined Pair Symbols:
LTE, EVEN
Compound Symbols:
c4, c8
0.00/0.85
0.00/0.85
(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
We considered the (Usable) Rules:none
And the Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
0.00/0.85
POL(Cons(x1, x2)) = [1] + x2
0.00/0.85
POL(EVEN(x1)) = 0
0.00/0.85
POL(LTE(x1, x2)) = x1·x2
0.00/0.85
POL(c4(x1)) = x1
0.00/0.85
POL(c8(x1)) = x1
0.00/0.85
0.00/0.85
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3)
0.00/0.85
lte(Cons(z0, z1), Nil) → False
0.00/0.85
lte(Nil, z0) → True
0.00/0.85
even(Cons(z0, Nil)) → False
0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2)
0.00/0.85
even(Nil) → True
0.00/0.85
notEmpty(Cons(z0, z1)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
S tuples:
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
K tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
Defined Rule Symbols:
lte, even, notEmpty, goal, and
Defined Pair Symbols:
LTE, EVEN
Compound Symbols:
c4, c8
0.00/0.85
0.00/0.85
(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
We considered the (Usable) Rules:none
And the Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
The order we found is given by the following interpretation:
Polynomial interpretation :
0.00/0.85
POL(Cons(x1, x2)) = [1] + x2
0.00/0.85
POL(EVEN(x1)) = [2]x1
0.00/0.85
POL(LTE(x1, x2)) = [3]x1 + [5]x2
0.00/0.85
POL(c4(x1)) = x1
0.00/0.85
POL(c8(x1)) = x1
0.00/0.85
0.00/0.85
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(False, False) → False
0.00/0.85
and(True, False) → False
0.00/0.85
and(False, True) → False
0.00/0.85
and(True, True) → True
0.00/0.85
lte(Cons(z0, z1), Cons(z2, z3)) → lte(z1, z3)
0.00/0.85
lte(Cons(z0, z1), Nil) → False
0.00/0.85
lte(Nil, z0) → True
0.00/0.85
even(Cons(z0, Nil)) → False
0.00/0.85
even(Cons(z0, Cons(z1, z2))) → even(z2)
0.00/0.85
even(Nil) → True
0.00/0.85
notEmpty(Cons(z0, z1)) → True
0.00/0.85
notEmpty(Nil) → False
0.00/0.85
goal(z0, z1) → and(lte(z0, z1), even(z0))
Tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
S tuples:none
K tuples:
LTE(Cons(z0, z1), Cons(z2, z3)) → c4(LTE(z1, z3))
0.00/0.85
EVEN(Cons(z0, Cons(z1, z2))) → c8(EVEN(z2))
Defined Rule Symbols:
lte, even, notEmpty, goal, and
Defined Pair Symbols:
LTE, EVEN
Compound Symbols:
c4, c8
0.00/0.85
0.00/0.85
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
0.00/0.85
0.00/0.85
(14) BOUNDS(O(1), O(1))
0.00/0.85