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