(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))
S tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(G(z0, if(f(z0), c(g(s(z0), z1)), c(z1))), IF(f(z0), c(g(s(z0), z1)), c(z1)), F(z0), G(s(z0), z1))
K tuples:none
Defined Rule Symbols:
f, if, g
Defined Pair Symbols:
F, G
Compound Symbols:
c3, c6, c7
(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
S tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
K tuples:none
Defined Rule Symbols:
f, if, g
Defined Pair Symbols:
F, G
Compound Symbols:
c3, c6, c7
(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.
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = [4]
POL(G(x1, x2)) = [4]x1 + [2]x2
POL(c(x1)) = [4] + x1
POL(c3(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(s(x1)) = 0
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
S tuples:
F(s(z0)) → c3(F(z0))
K tuples:
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
Defined Rule Symbols:
f, if, g
Defined Pair Symbols:
F, G
Compound Symbols:
c3, c6, c7
(7) 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(s(z0)) → c3(F(z0))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = x1
POL(G(x1, x2)) = x22 + x1·x2
POL(c(x1)) = [1] + x1
POL(c3(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1, x2)) = x1 + x2
POL(s(x1)) = [1] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → true
f(1) → false
f(s(z0)) → f(z0)
if(true, s(z0), s(z1)) → s(z0)
if(false, s(z0), s(z1)) → s(z1)
g(z0, c(z1)) → c(g(z0, z1))
g(z0, c(z1)) → g(z0, if(f(z0), c(g(s(z0), z1)), c(z1)))
Tuples:
F(s(z0)) → c3(F(z0))
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
S tuples:none
K tuples:
G(z0, c(z1)) → c6(G(z0, z1))
G(z0, c(z1)) → c7(F(z0), G(s(z0), z1))
F(s(z0)) → c3(F(z0))
Defined Rule Symbols:
f, if, g
Defined Pair Symbols:
F, G
Compound Symbols:
c3, c6, c7
(9) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(10) BOUNDS(O(1), O(1))