(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → minus(p(x), y)
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(0) → 0
p(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → minus(p(z0), z1)
Tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1), P(z0))
S tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1), P(z0))
K tuples:none
Defined Rule Symbols:
p, minus
Defined Pair Symbols:
MINUS
Compound Symbols:
c3
(3) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(0) → 0
p(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → minus(p(z0), z1)
Tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1))
S tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1))
K tuples:none
Defined Rule Symbols:
p, minus
Defined Pair Symbols:
MINUS
Compound Symbols:
c3
(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.
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1))
We considered the (Usable) Rules:
p(0) → 0
p(s(z0)) → z0
And the Tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [4]
POL(MINUS(x1, x2)) = [4]x2
POL(c3(x1)) = x1
POL(p(x1)) = [5]
POL(s(x1)) = [2] + x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(0) → 0
p(s(z0)) → z0
minus(z0, 0) → z0
minus(z0, s(z1)) → minus(p(z0), z1)
Tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1))
S tuples:none
K tuples:
MINUS(z0, s(z1)) → c3(MINUS(p(z0), z1))
Defined Rule Symbols:
p, minus
Defined Pair Symbols:
MINUS
Compound Symbols:
c3
(7) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(8) BOUNDS(O(1), O(1))