YES(O(1), O(n^1)) 0.00/0.73 YES(O(1), O(n^1)) 0.00/0.75 0.00/0.75 0.00/0.75
0.00/0.75 0.00/0.750 CpxTRS0.00/0.75
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))0.00/0.75
↳2 CdtProblem0.00/0.75
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))0.00/0.75
↳4 CdtProblem0.00/0.75
↳5 CdtKnowledgeProof (⇔)0.00/0.75
↳6 BOUNDS(O(1), O(1))0.00/0.75
f(0, 1, x) → f(s(x), x, x) 0.00/0.75
f(x, y, s(z)) → s(f(0, 1, z))
Tuples:
f(0, 1, z0) → f(s(z0), z0, z0) 0.00/0.75
f(z0, z1, s(z2)) → s(f(0, 1, z2))
S tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0)) 0.00/0.75
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
K tuples:none
F(0, 1, z0) → c(F(s(z0), z0, z0)) 0.00/0.75
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
f
F
c, c1
We considered the (Usable) Rules:none
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
The order we found is given by the following interpretation:
F(0, 1, z0) → c(F(s(z0), z0, z0)) 0.00/0.75
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
POL(0) = [1] 0.00/0.75
POL(1) = 0 0.00/0.75
POL(F(x1, x2, x3)) = [4]x3 0.00/0.75
POL(c(x1)) = x1 0.00/0.75
POL(c1(x1)) = x1 0.00/0.75
POL(s(x1)) = [2] + x1
Tuples:
f(0, 1, z0) → f(s(z0), z0, z0) 0.00/0.75
f(z0, z1, s(z2)) → s(f(0, 1, z2))
S tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0)) 0.00/0.75
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
K tuples:
F(0, 1, z0) → c(F(s(z0), z0, z0))
Defined Rule Symbols:
F(z0, z1, s(z2)) → c1(F(0, 1, z2))
f
F
c, c1
Now S is empty
F(0, 1, z0) → c(F(s(z0), z0, z0)) 0.00/0.75
F(z0, z1, s(z2)) → c1(F(0, 1, z2))