(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(b(c(z0))), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(b(c(z0))), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
K tuples:none
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A
Compound Symbols:
c1, c2, c3
(3) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
A(0, z0) → c3(C(c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
K tuples:none
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A
Compound Symbols:
c1, c3, c2
(5) CdtInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use instantiation to replace
A(
0,
z0) →
c3(
C(
c(
z0)),
C(
z0)) by
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c3(C(c(z0)), C(z0))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c3(C(c(z0)), C(z0))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c3(C(c(z0)), C(z0))
K tuples:none
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A, A1
Compound Symbols:
c1, c2, c3
(7) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)
Split RHS of tuples not part of any SCC
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
K tuples:none
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A, A1
Compound Symbols:
c1, c2, c3, c4
(9) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(x0)) → c3(C(c(c(x0))), C(c(x0)))
K tuples:
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A, A1
Compound Symbols:
c1, c2, c3, c4
(11) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
A(
0,
c(
x0)) →
c3(
C(
c(
c(
x0))),
C(
c(
x0))) by
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
K tuples:
A1(0, z0) → c4(C(c(z0)))
A1(0, z0) → c4(C(z0))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A1, A
Compound Symbols:
c1, c2, c4, c3
(13) CdtGraphRemoveDanglingProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 of 6 dangling nodes:
A1(0, z0) → c4(C(z0))
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(b(a(0, c(z0)))), C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(b(c(b(c(z0))))), C(c(z0)))
K tuples:
A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A1, A
Compound Symbols:
c1, c2, c4, c3
(15) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
S tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
K tuples:
A1(0, z0) → c4(C(c(z0)))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A1, A
Compound Symbols:
c1, c2, c4, c3
(17) 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.
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
We considered the (Usable) Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
And the Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(A(x1, x2)) = [1] + [5]x1 + x2
POL(A1(x1, x2)) = [4] + [4]x1 + [5]x2
POL(C(x1)) = x1
POL(a(x1, x2)) = [3] + [2]x1 + [4]x2
POL(b(x1)) = x1
POL(c(x1)) = [1] + [2]x1
POL(c1(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
(18) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
S tuples:
C(c(z0)) → c2(C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
K tuples:
A1(0, z0) → c4(C(c(z0)))
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A1, A
Compound Symbols:
c1, c2, c4, c3
(19) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
(20) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
S tuples:
C(c(z0)) → c2(C(z0))
K tuples:
A1(0, z0) → c4(C(c(z0)))
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A1, A
Compound Symbols:
c1, c2, c4, c3
(21) 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.
C(c(z0)) → c2(C(z0))
We considered the (Usable) Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
And the Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(A(x1, x2)) = [1] + [4]x1 + [2]x2
POL(A1(x1, x2)) = [3] + [4]x1 + [4]x2
POL(C(x1)) = [1] + [2]x1
POL(a(x1, x2)) = [3] + [4]x2
POL(b(x1)) = x1
POL(c(x1)) = [1] + [2]x1
POL(c1(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c4(x1)) = x1
(22) Obligation:
Complexity Dependency Tuples Problem
Rules:
c(c(b(c(z0)))) → b(a(0, c(z0)))
c(c(z0)) → b(c(b(c(z0))))
a(0, z0) → c(c(z0))
Tuples:
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
C(c(z0)) → c2(C(z0))
A1(0, z0) → c4(C(c(z0)))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
S tuples:none
K tuples:
A1(0, z0) → c4(C(c(z0)))
C(c(b(c(z0)))) → c1(A(0, c(z0)), C(z0))
A(0, c(b(c(z0)))) → c3(C(c(b(c(z0)))))
A(0, c(z0)) → c3(C(c(z0)))
C(c(z0)) → c2(C(z0))
Defined Rule Symbols:
c, a
Defined Pair Symbols:
C, A1, A
Compound Symbols:
c1, c2, c4, c3
(23) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(24) BOUNDS(O(1), O(1))