(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(c(n, x)) → c(f(g(c(n, x))), f(h(c(n, x))))
g(c(0, x)) → x
h(c(1, x)) → x
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(z0, z1)) → c1(F(g(c(z0, z1))), G(c(z0, z1)), F(h(c(z0, z1))), H(c(z0, z1)))
S tuples:
F(c(z0, z1)) → c1(F(g(c(z0, z1))), G(c(z0, z1)), F(h(c(z0, z1))), H(c(z0, z1)))
K tuples:none
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1
(3) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
F(
c(
z0,
z1)) →
c1(
F(
g(
c(
z0,
z1))),
G(
c(
z0,
z1)),
F(
h(
c(
z0,
z1))),
H(
c(
z0,
z1))) by
F(c(0, z0)) → c1(F(z0), G(c(0, z0)), F(h(c(0, z0))), H(c(0, z0)))
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(0, z0)) → c1(F(z0), G(c(0, z0)), F(h(c(0, z0))), H(c(0, z0)))
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
S tuples:
F(c(0, z0)) → c1(F(z0), G(c(0, z0)), F(h(c(0, z0))), H(c(0, z0)))
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
K tuples:none
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1, c1
(5) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
F(
c(
0,
z0)) →
c1(
F(
z0),
G(
c(
0,
z0)),
F(
h(
c(
0,
z0))),
H(
c(
0,
z0))) by
F(c(0, x0)) → c1(F(x0))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, x0)) → c1(F(x0))
S tuples:
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, x0)) → c1(F(x0))
K tuples:none
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1
(7) 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.
F(c(0, x0)) → c1(F(x0))
We considered the (Usable) Rules:
h(c(1, z0)) → z0
And the Tuples:
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, x0)) → c1(F(x0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [5]
POL(1) = 0
POL(F(x1)) = [2]x1
POL(c(x1, x2)) = [1] + x2
POL(c1(x1)) = x1
POL(h(x1)) = x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
F(c(0, x0)) → c1(F(x0))
S tuples:
F(c(x0, x1)) → c1(F(h(c(x0, x1))))
K tuples:
F(c(0, x0)) → c1(F(x0))
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1
(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
F(
c(
x0,
x1)) →
c1(
F(
h(
c(
x0,
x1)))) by
F(c(1, z0)) → c1(F(z0))
F(c(x0, x1)) → c1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(0, x0)) → c1(F(x0))
F(c(1, z0)) → c1(F(z0))
F(c(x0, x1)) → c1
S tuples:
F(c(1, z0)) → c1(F(z0))
F(c(x0, x1)) → c1
K tuples:
F(c(0, x0)) → c1(F(x0))
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1, c1
(11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
F(c(x0, x1)) → c1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(0, x0)) → c1(F(x0))
F(c(1, z0)) → c1(F(z0))
S tuples:
F(c(1, z0)) → c1(F(z0))
K tuples:
F(c(0, x0)) → c1(F(x0))
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1
(13) 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.
F(c(1, z0)) → c1(F(z0))
We considered the (Usable) Rules:none
And the Tuples:
F(c(0, x0)) → c1(F(x0))
F(c(1, z0)) → c1(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(1) = [1]
POL(F(x1)) = x12
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(c(z0, z1)) → c(f(g(c(z0, z1))), f(h(c(z0, z1))))
g(c(0, z0)) → z0
h(c(1, z0)) → z0
Tuples:
F(c(0, x0)) → c1(F(x0))
F(c(1, z0)) → c1(F(z0))
S tuples:none
K tuples:
F(c(0, x0)) → c1(F(x0))
F(c(1, z0)) → c1(F(z0))
Defined Rule Symbols:
f, g, h
Defined Pair Symbols:
F
Compound Symbols:
c1
(15) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(16) BOUNDS(O(1), O(1))