YES(O(1), O(n^1)) 0.00/0.82 YES(O(1), O(n^1)) 0.00/0.83 0.00/0.83 0.00/0.83
0.00/0.83 0.00/0.830 CpxTRS0.00/0.83
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))0.00/0.83
↳2 CdtProblem0.00/0.83
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))0.00/0.83
↳4 CdtProblem0.00/0.83
↳5 CdtNarrowingProof (BOTH BOUNDS(ID, ID))0.00/0.83
↳6 CdtProblem0.00/0.83
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))0.00/0.83
↳8 CdtProblem0.00/0.83
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))0.00/0.83
↳10 BOUNDS(O(1), O(1))0.00/0.83
f(s(x)) → s(s(f(p(s(x))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(x)) → x
Tuples:
f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
S tuples:
F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
K tuples:none
F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
f, p
F
c
Tuples:
f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
S tuples:
F(s(z0)) → c(F(p(s(z0))))
K tuples:none
F(s(z0)) → c(F(p(s(z0))))
f, p
F
c
F(s(z0)) → c(F(z0))
Tuples:
f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
S tuples:
F(s(z0)) → c(F(z0))
K tuples:none
F(s(z0)) → c(F(z0))
f, p
F
c
We considered the (Usable) Rules:none
F(s(z0)) → c(F(z0))
The order we found is given by the following interpretation:
F(s(z0)) → c(F(z0))
POL(F(x1)) = [3]x1 0.00/0.83
POL(c(x1)) = x1 0.00/0.83
POL(s(x1)) = [1] + x1
Tuples:
f(s(z0)) → s(s(f(p(s(z0))))) 0.00/0.83
f(0) → 0 0.00/0.83
p(s(z0)) → z0
S tuples:none
F(s(z0)) → c(F(z0))
Defined Rule Symbols:
F(s(z0)) → c(F(z0))
f, p
F
c