(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(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.
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
We considered the (Usable) Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
And the Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [4]
POL(F(x1, x2)) = [3]
POL(G(x1, x2)) = [4]x1
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(f(x1, x2)) = 0
POL(s(x1)) = 0
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(5) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
G(
0,
z0) →
c2(
G(
f(
z0,
z0),
z0),
F(
z0,
z0)) by
G(0, 0) → c2(G(s(0), 0), F(0, 0))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, 0) → c2(G(s(0), 0), F(0, 0))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:
G(0, z0) → c2(G(f(z0, z0), z0), F(z0, z0))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(7) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)
Removed 1 of 3 dangling nodes:
G(0, 0) → c2(G(s(0), 0), F(0, 0))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c2(G(s(f(z0, z0)), s(z0)), F(s(z0), s(z0)))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c2
(9) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)
Split RHS of tuples not part of any SCC
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c(G(s(f(z0, z0)), s(z0)))
G(0, s(z0)) → c(F(s(z0), s(z0)))
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c
(11) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c(F(s(z0), s(z0)))
G(0, s(z0)) → c
S tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
K tuples:none
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c, c
(13) 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(s(z0), s(z1)) → c1(F(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c(F(s(z0), s(z0)))
G(0, s(z0)) → c
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [2]
POL(F(x1, x2)) = [3] + x2
POL(G(x1, x2)) = [2] + x1 + [2]x2
POL(c) = 0
POL(c(x1)) = x1
POL(c1(x1)) = x1
POL(s(x1)) = [2] + x1
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(z0, 0) → s(0)
f(s(z0), s(z1)) → s(f(z0, z1))
g(0, z0) → g(f(z0, z0), z0)
Tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
G(0, s(z0)) → c(F(s(z0), s(z0)))
G(0, s(z0)) → c
S tuples:none
K tuples:
F(s(z0), s(z1)) → c1(F(z0, z1))
Defined Rule Symbols:
f, g
Defined Pair Symbols:
F, G
Compound Symbols:
c1, c, c
(15) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(16) BOUNDS(O(1), O(1))