(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
S tuples:
-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
K tuples:none
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(3) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
-'(
z0,
s(
z1)) →
c2(
-'(
z0,
p(
s(
z1))),
P(
s(
z1))) by
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
-'(x0, s(x1)) → c2
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
-'(x0, s(x1)) → c2
S tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
-'(x0, s(x1)) → c2
K tuples:none
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-'
Compound Symbols:
c2, c2
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
-'(x0, s(x1)) → c2
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
S tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
K tuples:none
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(7) 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.
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
We considered the (Usable) Rules:none
And the Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(-'(x1, x2)) = [4]x2
POL(P(x1)) = 0
POL(c2(x1, x2)) = x1 + x2
POL(s(x1)) = [4] + x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
-(0, z0) → 0
-(z0, 0) → z0
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0)
p(0) → 0
p(s(z0)) → z0
Tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
S tuples:none
K tuples:
-'(x0, s(z0)) → c2(-'(x0, z0), P(s(z0)))
Defined Rule Symbols:
-, p
Defined Pair Symbols:
-'
Compound Symbols:
c2
(9) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(10) BOUNDS(O(1), O(1))