YES(O(1), O(n^1)) 0.00/0.71 YES(O(1), O(n^1)) 0.00/0.72 0.00/0.72 0.00/0.72
0.00/0.72 0.00/0.720 CpxTRS0.00/0.72
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))0.00/0.72
↳2 CdtProblem0.00/0.72
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))0.00/0.72
↳4 CdtProblem0.00/0.72
↳5 SIsEmptyProof (BOTH BOUNDS(ID, ID))0.00/0.72
↳6 BOUNDS(O(1), O(1))0.00/0.72
g(s(x)) → f(x) 0.00/0.72
f(0) → s(0) 0.00/0.72
f(s(x)) → s(s(g(x))) 0.00/0.72
g(0) → 0
Tuples:
g(s(z0)) → f(z0) 0.00/0.72
g(0) → 0 0.00/0.72
f(0) → s(0) 0.00/0.72
f(s(z0)) → s(s(g(z0)))
S tuples:
G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
K tuples:none
G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
g, f
G, F
c, c3
We considered the (Usable) Rules:none
G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
The order we found is given by the following interpretation:
G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
POL(F(x1)) = [2] + [2]x1 0.00/0.72
POL(G(x1)) = [4] + [2]x1 0.00/0.72
POL(c(x1)) = x1 0.00/0.72
POL(c3(x1)) = x1 0.00/0.72
POL(s(x1)) = [4] + x1
Tuples:
g(s(z0)) → f(z0) 0.00/0.72
g(0) → 0 0.00/0.72
f(0) → s(0) 0.00/0.72
f(s(z0)) → s(s(g(z0)))
S tuples:none
G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
Defined Rule Symbols:
G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
g, f
G, F
c, c3