(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(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.
F(c, c) → c4(F(a, a))
We considered the (Usable) Rules:none
And the Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x1
POL(a) = 0
POL(b) = [5]
POL(c) = [2]
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
POL(s(x1)) = x1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
K tuples:
F(c, c) → c4(F(a, a))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(a, b) → c2(F(s(a), c))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(s(z0), c) → c3(F(z0, c))
F(c, c) → c4(F(a, a))
S tuples:
F(s(z0), c) → c3(F(z0, c))
K tuples:
F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c3, c4
(7) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use forward instantiation to replace
F(
s(
z0),
c) →
c3(
F(
z0,
c)) by
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(c, c) → c4(F(a, a))
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))
S tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
F(s(c), c) → c3(F(c, c))
K tuples:
F(c, c) → c4(F(a, a))
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c1, c2, c4, c3
(9) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)
Removed 4 of 5 dangling nodes:
F(a, a) → c1(F(a, b))
F(a, b) → c2(F(s(a), c))
F(c, c) → c4(F(a, a))
F(s(c), c) → c3(F(c, c))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
S tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c3
(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.
F(s(s(y0)), c) → c3(F(s(y0), c))
We considered the (Usable) Rules:none
And the Tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1, x2)) = x1
POL(c) = 0
POL(c3(x1)) = x1
POL(s(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(z0), c) → f(z0, c)
f(c, c) → f(a, a)
Tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
S tuples:none
K tuples:
F(s(s(y0)), c) → c3(F(s(y0), c))
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c3
(13) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(14) BOUNDS(O(1), O(1))