(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0
Tuples:
AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:
AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:
and, plus, activate
Defined Pair Symbols:
AND, PLUS
Compound Symbols:
c, c2
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
AND(tt, z0) → c(ACTIVATE(z0))
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0
Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:
and, plus, activate
Defined Pair Symbols:
PLUS
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.
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(PLUS(x1, x2)) = [5]x2
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0
Tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:none
K tuples:
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
Defined Rule Symbols:
and, plus, activate
Defined Pair Symbols:
PLUS
Compound Symbols:
c2
(7) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(8) BOUNDS(O(1), O(1))