We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict Trs:
{ f(x, y, z) -> g(<=(x, y), x, y, z)
, g(true(), x, y, z) -> z
, g(false(), x, y, z) ->
f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y))
, p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
We add the following dependency tuples:
Strict DPs:
{ f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, g^#(true(), x, y, z) -> c_2()
, g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z))
, p^#(0()) -> c_4()
, p^#(s(x)) -> c_5() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, g^#(true(), x, y, z) -> c_2()
, g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z))
, p^#(0()) -> c_4()
, p^#(s(x)) -> c_5() }
Weak Trs:
{ f(x, y, z) -> g(<=(x, y), x, y, z)
, g(true(), x, y, z) -> z
, g(false(), x, y, z) ->
f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y))
, p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
We estimate the number of application of {1,2,4,5} by applications
of Pre({1,2,4,5}) = {3}. Here rules are labeled as follows:
DPs:
{ 1: f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, 2: g^#(true(), x, y, z) -> c_2()
, 3: g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z))
, 4: p^#(0()) -> c_4()
, 5: p^#(s(x)) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z)) }
Weak DPs:
{ f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, g^#(true(), x, y, z) -> c_2()
, p^#(0()) -> c_4()
, p^#(s(x)) -> c_5() }
Weak Trs:
{ f(x, y, z) -> g(<=(x, y), x, y, z)
, g(true(), x, y, z) -> z
, g(false(), x, y, z) ->
f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y))
, p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
We estimate the number of application of {1} by applications of
Pre({1}) = {}. Here rules are labeled as follows:
DPs:
{ 1: g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z))
, 2: f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, 3: g^#(true(), x, y, z) -> c_2()
, 4: p^#(0()) -> c_4()
, 5: p^#(s(x)) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, g^#(true(), x, y, z) -> c_2()
, g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z))
, p^#(0()) -> c_4()
, p^#(s(x)) -> c_5() }
Weak Trs:
{ f(x, y, z) -> g(<=(x, y), x, y, z)
, g(true(), x, y, z) -> z
, g(false(), x, y, z) ->
f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y))
, p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ f^#(x, y, z) -> c_1(g^#(<=(x, y), x, y, z))
, g^#(true(), x, y, z) -> c_2()
, g^#(false(), x, y, z) ->
c_3(f^#(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y)),
f^#(p(x), y, z),
p^#(x),
f^#(p(y), z, x),
p^#(y),
f^#(p(z), x, y),
p^#(z))
, p^#(0()) -> c_4()
, p^#(s(x)) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ f(x, y, z) -> g(<=(x, y), x, y, z)
, g(true(), x, y, z) -> z
, g(false(), x, y, z) ->
f(f(p(x), y, z), f(p(y), z, x), f(p(z), x, y))
, p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
No rule is usable, rules are removed from the input problem.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(1))