(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
q(0, s(y), s(z)) → 0
q(s(x), s(y), z) → q(x, y, z)
q(x, 0, s(z)) → s(q(x, s(z), s(z)))
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
q(0, s(z0), s(z1)) → 0
q(s(z0), s(z1), z2) → q(z0, z1, z2)
q(z0, 0, s(z1)) → s(q(z0, s(z1), s(z1)))
Tuples:
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
S tuples:
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:
q
Defined Pair Symbols:
Q
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.
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(Q(x1, x2, x3)) = [5]x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
q(0, s(z0), s(z1)) → 0
q(s(z0), s(z1), z2) → q(z0, z1, z2)
q(z0, 0, s(z1)) → s(q(z0, s(z1), s(z1)))
Tuples:
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
S tuples:
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
K tuples:
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Defined Rule Symbols:
q
Defined Pair Symbols:
Q
Compound Symbols:
c1, c2
(5) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
Q(z0, 0, s(z1)) → c2(Q(z0, s(z1), s(z1)))
Q(s(z0), s(z1), z2) → c1(Q(z0, z1, z2))
Now S is empty
(6) BOUNDS(O(1), O(1))