(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
D(minus(z0)) → minus(D(z0))
D(div(z0, z1)) → -(div(D(z0), z1), div(*(z0, D(z1)), pow(z1, 2)))
D(ln(z0)) → div(D(z0), z0)
D(pow(z0, z1)) → +(*(*(z1, pow(z0, -(z1, 1))), D(z0)), *(*(pow(z0, z1), ln(z0)), D(z1)))
Tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
S tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:
D
Defined Pair Symbols:
D'
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8
(3) 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.
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(ln(z0)) → c7(D'(z0))
We considered the (Usable) Rules:none
And the Tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*(x1, x2)) = [3] + x1 + x2
POL(+(x1, x2)) = [4] + x1 + x2
POL(-(x1, x2)) = [2] + x1 + x2
POL(D'(x1)) = [2]x1
POL(c2(x1, x2)) = x1 + x2
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(div(x1, x2)) = x1 + x2
POL(ln(x1)) = [3] + x1
POL(minus(x1)) = [5] + x1
POL(pow(x1, x2)) = x1 + x2
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
D(minus(z0)) → minus(D(z0))
D(div(z0, z1)) → -(div(D(z0), z1), div(*(z0, D(z1)), pow(z1, 2)))
D(ln(z0)) → div(D(z0), z0)
D(pow(z0, z1)) → +(*(*(z1, pow(z0, -(z1, 1))), D(z0)), *(*(pow(z0, z1), ln(z0)), D(z1)))
Tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
S tuples:
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
K tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(ln(z0)) → c7(D'(z0))
Defined Rule Symbols:
D
Defined Pair Symbols:
D'
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8
(5) 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.
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
We considered the (Usable) Rules:none
And the Tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(*(x1, x2)) = [2] + x1 + x2
POL(+(x1, x2)) = [4] + x1 + x2
POL(-(x1, x2)) = [5] + x1 + x2
POL(D'(x1)) = [2] + [4]x1
POL(c2(x1, x2)) = x1 + x2
POL(c3(x1, x2)) = x1 + x2
POL(c4(x1, x2)) = x1 + x2
POL(c5(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(div(x1, x2)) = [5] + x1 + x2
POL(ln(x1)) = x1
POL(minus(x1)) = x1
POL(pow(x1, x2)) = [4] + x1 + x2
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
D(t) → 1
D(constant) → 0
D(+(z0, z1)) → +(D(z0), D(z1))
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1)))
D(-(z0, z1)) → -(D(z0), D(z1))
D(minus(z0)) → minus(D(z0))
D(div(z0, z1)) → -(div(D(z0), z1), div(*(z0, D(z1)), pow(z1, 2)))
D(ln(z0)) → div(D(z0), z0)
D(pow(z0, z1)) → +(*(*(z1, pow(z0, -(z1, 1))), D(z0)), *(*(pow(z0, z1), ln(z0)), D(z1)))
Tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(ln(z0)) → c7(D'(z0))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
S tuples:none
K tuples:
D'(+(z0, z1)) → c2(D'(z0), D'(z1))
D'(*(z0, z1)) → c3(D'(z0), D'(z1))
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
D'(minus(z0)) → c5(D'(z0))
D'(ln(z0)) → c7(D'(z0))
D'(div(z0, z1)) → c6(D'(z0), D'(z1))
D'(pow(z0, z1)) → c8(D'(z0), D'(z1))
Defined Rule Symbols:
D
Defined Pair Symbols:
D'
Compound Symbols:
c2, c3, c4, c5, c6, c7, c8
(7) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(8) BOUNDS(O(1), O(1))