(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
P(s(0)) → c3(0')
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')
S tuples:
F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
P(s(0)) → c3(0')
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
F, P, ACTIVATE
Compound Symbols:
c, c1, c3, c6, c7, c8
(3) CdtUnreachableProof (EQUIVALENT transformation)
The following tuples could be removed as they are not reachable from basic start terms:
F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
P(s(0)) → c3(0')
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')
S tuples:
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7, c8
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
ACTIVATE(n__0) → c8(0')
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(9) 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.
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(ACTIVATE(x1)) = [4]x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(n__f(x1)) = [1] + x1
POL(n__s(x1)) = [4] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:none
K tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(11) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(12) BOUNDS(O(1), O(1))