We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
, plus(N, 0()) -> U11(isNat(N), N)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
, x(N, 0()) -> U31(isNat(N))
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
, plus(N, 0()) -> U11(isNat(N), N)
, x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
, x(N, 0()) -> U31(isNat(N)) }
All above mentioned rules can be savely removed.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following dependency tuples:
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(X) -> c_2()
, activate^#(n__0()) -> c_3(0^#())
, activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, activate^#(n__s(X)) -> c_6(s^#(X))
, activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, 0^#() -> c_12()
, plus^#(X1, X2) -> c_10()
, isNat^#(X) -> c_16()
, isNat^#(n__0()) -> c_17()
, isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, s^#(X) -> c_9()
, x^#(X1, X2) -> c_14()
, U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, U31^#(tt()) -> c_11(0^#())
, U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, and^#(tt(), X) -> c_15(activate^#(X)) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(X) -> c_2()
, activate^#(n__0()) -> c_3(0^#())
, activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, activate^#(n__s(X)) -> c_6(s^#(X))
, activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, 0^#() -> c_12()
, plus^#(X1, X2) -> c_10()
, isNat^#(X) -> c_16()
, isNat^#(n__0()) -> c_17()
, isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, s^#(X) -> c_9()
, x^#(X1, X2) -> c_14()
, U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, U31^#(tt()) -> c_11(0^#())
, U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, and^#(tt(), X) -> c_15(activate^#(X)) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {2,8,9,10,11,15,16} by
applications of Pre({2,8,9,10,11,15,16}) =
{1,3,4,5,6,7,12,13,14,17,18,19,20}. Here rules are labeled as
follows:
DPs:
{ 1: U11^#(tt(), N) -> c_1(activate^#(N))
, 2: activate^#(X) -> c_2()
, 3: activate^#(n__0()) -> c_3(0^#())
, 4: activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, 5: activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, 6: activate^#(n__s(X)) -> c_6(s^#(X))
, 7: activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, 8: 0^#() -> c_12()
, 9: plus^#(X1, X2) -> c_10()
, 10: isNat^#(X) -> c_16()
, 11: isNat^#(n__0()) -> c_17()
, 12: isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 13: isNat^#(n__s(V1)) ->
c_19(isNat^#(activate(V1)), activate^#(V1))
, 14: isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 15: s^#(X) -> c_9()
, 16: x^#(X1, X2) -> c_14()
, 17: U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, 18: U31^#(tt()) -> c_11(0^#())
, 19: U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, 20: and^#(tt(), X) -> c_15(activate^#(X)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(n__0()) -> c_3(0^#())
, activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, activate^#(n__s(X)) -> c_6(s^#(X))
, activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, U31^#(tt()) -> c_11(0^#())
, U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, and^#(tt(), X) -> c_15(activate^#(X)) }
Weak DPs:
{ activate^#(X) -> c_2()
, 0^#() -> c_12()
, plus^#(X1, X2) -> c_10()
, isNat^#(X) -> c_16()
, isNat^#(n__0()) -> c_17()
, s^#(X) -> c_9()
, x^#(X1, X2) -> c_14() }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {2,3,5,6,11} by
applications of Pre({2,3,5,6,11}) = {1,7,8,9,10,12,13}. Here rules
are labeled as follows:
DPs:
{ 1: U11^#(tt(), N) -> c_1(activate^#(N))
, 2: activate^#(n__0()) -> c_3(0^#())
, 3: activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, 4: activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, 5: activate^#(n__s(X)) -> c_6(s^#(X))
, 6: activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, 7: isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 8: isNat^#(n__s(V1)) ->
c_19(isNat^#(activate(V1)), activate^#(V1))
, 9: isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 10: U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, 11: U31^#(tt()) -> c_11(0^#())
, 12: U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, 13: and^#(tt(), X) -> c_15(activate^#(X))
, 14: activate^#(X) -> c_2()
, 15: 0^#() -> c_12()
, 16: plus^#(X1, X2) -> c_10()
, 17: isNat^#(X) -> c_16()
, 18: isNat^#(n__0()) -> c_17()
, 19: s^#(X) -> c_9()
, 20: x^#(X1, X2) -> c_14() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, and^#(tt(), X) -> c_15(activate^#(X)) }
Weak DPs:
{ activate^#(X) -> c_2()
, activate^#(n__0()) -> c_3(0^#())
, activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, activate^#(n__s(X)) -> c_6(s^#(X))
, activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, 0^#() -> c_12()
, plus^#(X1, X2) -> c_10()
, isNat^#(X) -> c_16()
, isNat^#(n__0()) -> c_17()
, s^#(X) -> c_9()
, x^#(X1, X2) -> c_14()
, U31^#(tt()) -> c_11(0^#()) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ activate^#(X) -> c_2()
, activate^#(n__0()) -> c_3(0^#())
, activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2))
, activate^#(n__s(X)) -> c_6(s^#(X))
, activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2))
, 0^#() -> c_12()
, plus^#(X1, X2) -> c_10()
, isNat^#(X) -> c_16()
, isNat^#(n__0()) -> c_17()
, s^#(X) -> c_9()
, x^#(X1, X2) -> c_14()
, U31^#(tt()) -> c_11(0^#()) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(n__isNat(X)) -> c_5(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N))
, and^#(tt(), X) -> c_15(activate^#(X)) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:
{ U21^#(tt(), M, N) ->
c_8(s^#(plus(activate(N), activate(M))),
plus^#(activate(N), activate(M)),
activate^#(N),
activate^#(M))
, U41^#(tt(), M, N) ->
c_13(plus^#(x(activate(N), activate(M)), activate(N)),
x^#(activate(N), activate(M)),
activate^#(N),
activate^#(M),
activate^#(N)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M))
, U41^#(tt(), M, N) ->
c_7(activate^#(N), activate^#(M), activate^#(N))
, and^#(tt(), X) -> c_8(activate^#(X)) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ U11^#(tt(), N) -> c_1(activate^#(N))
, activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M))
, U41^#(tt(), M, N) ->
c_7(activate^#(N), activate^#(M), activate^#(N))
, and^#(tt(), X) -> c_8(activate^#(X)) }
Weak Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Consider the dependency graph
1: U11^#(tt(), N) -> c_1(activate^#(N))
-->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
2: activate^#(n__isNat(X)) -> c_2(isNat^#(X))
-->_1 isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :5
-->_1 isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1)) :4
-->_1 isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :3
3: isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
-->_1 and^#(tt(), X) -> c_8(activate^#(X)) :8
-->_2 isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :5
-->_2 isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1)) :4
-->_2 isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :3
-->_4 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
-->_3 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
4: isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1))
-->_1 isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :5
-->_1 isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1)) :4
-->_1 isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :3
-->_2 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
5: isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
-->_1 and^#(tt(), X) -> c_8(activate^#(X)) :8
-->_2 isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :5
-->_2 isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1)) :4
-->_2 isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) :3
-->_4 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
-->_3 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
6: U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M))
-->_2 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
-->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
7: U41^#(tt(), M, N) ->
c_7(activate^#(N), activate^#(M), activate^#(N))
-->_3 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
-->_2 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
-->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
8: and^#(tt(), X) -> c_8(activate^#(X))
-->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2
Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).
{ U11^#(tt(), N) -> c_1(activate^#(N))
, U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M))
, U41^#(tt(), M, N) ->
c_7(activate^#(N), activate^#(M), activate^#(N)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, and^#(tt(), X) -> c_8(activate^#(X)) }
Weak Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
DPs:
{ 2: isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 4: isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) }
Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, isNat(X) -> n__isNat(X)
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {1}, Uargs(c_3) = {1, 2, 3, 4}, Uargs(c_4) = {1, 2},
Uargs(c_5) = {1, 2, 3, 4}, Uargs(c_8) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[tt] = [4]
[3]
[activate](x1) = [1 0] x1 + [4]
[0 1] [1]
[s](x1) = [0 0] x1 + [3]
[1 1] [2]
[plus](x1, x2) = [0 0] x1 + [0 0] x2 + [4]
[1 1] [1 1] [7]
[0] = [1]
[0]
[x](x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[1 1] [1 1] [4]
[and](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [1]
[isNat](x1) = [0 1] x1 + [4]
[0 0] [3]
[n__0] = [1]
[0]
[n__plus](x1, x2) = [0 0] x1 + [0 0] x2 + [4]
[1 1] [1 1] [7]
[n__isNat](x1) = [0 1] x1 + [1]
[0 0] [2]
[n__s](x1) = [0 0] x1 + [1]
[1 1] [2]
[n__x](x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[1 1] [1 1] [4]
[activate^#](x1) = [2 0] x1 + [0]
[0 0] [2]
[isNat^#](x1) = [0 2] x1 + [1]
[0 0] [0]
[and^#](x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 1] [0 0] [0]
[c_2](x1) = [1 1] x1 + [1]
[0 0] [0]
[c_3](x1, x2, x3, x4) = [1 1] x1 + [1 2] x2 + [1 0] x3 + [1
2] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[c_4](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 0] [0 0] [0]
[c_5](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1
0] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[c_8](x1) = [1 0] x1 + [0]
[0 0] [0]
The order satisfies the following ordering constraints:
[activate(X)] = [1 0] X + [4]
[0 1] [1]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [5]
[1]
> [1]
[0]
= [0()]
[activate(n__plus(X1, X2))] = [0 0] X1 + [0 0] X2 + [8]
[1 1] [1 1] [8]
> [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [7]
= [plus(X1, X2)]
[activate(n__isNat(X))] = [0 1] X + [5]
[0 0] [3]
> [0 1] X + [4]
[0 0] [3]
= [isNat(X)]
[activate(n__s(X))] = [0 0] X + [5]
[1 1] [3]
> [0 0] X + [3]
[1 1] [2]
= [s(X)]
[activate(n__x(X1, X2))] = [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [5]
> [0 0] X1 + [0 0] X2 + [0]
[1 1] [1 1] [4]
= [x(X1, X2)]
[s(X)] = [0 0] X + [3]
[1 1] [2]
> [0 0] X + [1]
[1 1] [2]
= [n__s(X)]
[plus(X1, X2)] = [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [7]
>= [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [7]
= [n__plus(X1, X2)]
[0()] = [1]
[0]
>= [1]
[0]
= [n__0()]
[x(X1, X2)] = [0 0] X1 + [0 0] X2 + [0]
[1 1] [1 1] [4]
>= [0 0] X1 + [0 0] X2 + [0]
[1 1] [1 1] [4]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 0] X + [4]
[0 1] [1]
>= [1 0] X + [4]
[0 1] [1]
= [activate(X)]
[isNat(X)] = [0 1] X + [4]
[0 0] [3]
> [0 1] X + [1]
[0 0] [2]
= [n__isNat(X)]
[isNat(n__0())] = [4]
[3]
>= [4]
[3]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [11]
[0 0] [0 0] [3]
> [0 1] V1 + [0 1] V2 + [7]
[0 0] [0 0] [3]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 1] V1 + [6]
[0 0] [3]
> [0 1] V1 + [5]
[0 0] [3]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 1] V1 + [1 1] V2 + [8]
[0 0] [0 0] [3]
> [0 1] V1 + [0 1] V2 + [7]
[0 0] [0 0] [3]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[activate^#(n__isNat(X))] = [0 2] X + [2]
[0 0] [2]
>= [0 2] X + [2]
[0 0] [0]
= [c_2(isNat^#(X))]
[isNat^#(n__plus(V1, V2))] = [2 2] V1 + [2 2] V2 + [15]
[0 0] [0 0] [0]
> [2 2] V1 + [2 2] V2 + [14]
[0 0] [0 0] [0]
= [c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))]
[isNat^#(n__s(V1))] = [2 2] V1 + [5]
[0 0] [0]
>= [2 2] V1 + [5]
[0 0] [0]
= [c_4(isNat^#(activate(V1)), activate^#(V1))]
[isNat^#(n__x(V1, V2))] = [2 2] V1 + [2 2] V2 + [9]
[0 0] [0 0] [0]
> [2 2] V1 + [2 2] V2 + [7]
[0 0] [0 0] [0]
= [c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))]
[and^#(tt(), X)] = [2 0] X + [0]
[0 0] [3]
>= [2 0] X + [0]
[0 0] [0]
= [c_8(activate^#(X))]
We return to the main proof. Consider the set of all dependency
pairs
:
{ 1: activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, 2: isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 3: isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1))
, 4: isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 5: and^#(tt(), X) -> c_8(activate^#(X)) }
Processor 'matrix interpretation of dimension 2' induces the
complexity certificate YES(?,O(n^1)) on application of dependency
pairs {2,4}. These cover all (indirect) predecessors of dependency
pairs {2,4,5}, their number of application is equally bounded. The
dependency pairs are shifted into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) }
Weak DPs:
{ isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, and^#(tt(), X) -> c_8(activate^#(X)) }
Weak Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
DPs:
{ 1: activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, 3: isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 4: isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) }
Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, 0() -> n__0()
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {1}, Uargs(c_3) = {1, 2, 3, 4}, Uargs(c_4) = {1, 2},
Uargs(c_5) = {1, 2, 3, 4}, Uargs(c_8) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[tt] = [0]
[0]
[activate](x1) = [1 0] x1 + [4]
[0 1] [0]
[s](x1) = [0 0] x1 + [1]
[1 1] [0]
[plus](x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[1 1] [1 1] [7]
[0] = [1]
[2]
[x](x1, x2) = [0 0] x1 + [0 0] x2 + [4]
[1 1] [1 1] [5]
[and](x1, x2) = [1 2] x2 + [4]
[0 1] [0]
[isNat](x1) = [0 1] x1 + [7]
[0 0] [2]
[n__0] = [0]
[2]
[n__plus](x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[1 1] [1 1] [7]
[n__isNat](x1) = [0 1] x1 + [4]
[0 0] [2]
[n__s](x1) = [0 0] x1 + [1]
[1 1] [0]
[n__x](x1, x2) = [0 0] x1 + [0 0] x2 + [4]
[1 1] [1 1] [5]
[activate^#](x1) = [2 0] x1 + [0]
[0 0] [0]
[isNat^#](x1) = [0 2] x1 + [1]
[0 1] [2]
[and^#](x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 1] [0 0] [1]
[c_2](x1) = [1 0] x1 + [6]
[0 0] [0]
[c_3](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1
0] x4 + [1]
[0 0] [0 0] [0 0] [0
0] [0]
[c_4](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 0] [0 0] [0]
[c_5](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1
0] x4 + [1]
[0 0] [0 0] [0 0] [0
0] [0]
[c_8](x1) = [1 0] x1 + [0]
[0 0] [0]
The order satisfies the following ordering constraints:
[activate(X)] = [1 0] X + [4]
[0 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [4]
[2]
> [1]
[2]
= [0()]
[activate(n__plus(X1, X2))] = [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [7]
> [0 0] X1 + [0 0] X2 + [0]
[1 1] [1 1] [7]
= [plus(X1, X2)]
[activate(n__isNat(X))] = [0 1] X + [8]
[0 0] [2]
> [0 1] X + [7]
[0 0] [2]
= [isNat(X)]
[activate(n__s(X))] = [0 0] X + [5]
[1 1] [0]
> [0 0] X + [1]
[1 1] [0]
= [s(X)]
[activate(n__x(X1, X2))] = [0 0] X1 + [0 0] X2 + [8]
[1 1] [1 1] [5]
> [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [5]
= [x(X1, X2)]
[s(X)] = [0 0] X + [1]
[1 1] [0]
>= [0 0] X + [1]
[1 1] [0]
= [n__s(X)]
[plus(X1, X2)] = [0 0] X1 + [0 0] X2 + [0]
[1 1] [1 1] [7]
>= [0 0] X1 + [0 0] X2 + [0]
[1 1] [1 1] [7]
= [n__plus(X1, X2)]
[0()] = [1]
[2]
> [0]
[2]
= [n__0()]
[x(X1, X2)] = [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [5]
>= [0 0] X1 + [0 0] X2 + [4]
[1 1] [1 1] [5]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 2] X + [4]
[0 1] [0]
>= [1 0] X + [4]
[0 1] [0]
= [activate(X)]
[isNat(X)] = [0 1] X + [7]
[0 0] [2]
> [0 1] X + [4]
[0 0] [2]
= [n__isNat(X)]
[isNat(n__0())] = [9]
[2]
> [0]
[0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [14]
[0 0] [0 0] [2]
> [0 1] V2 + [12]
[0 0] [2]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 1] V1 + [7]
[0 0] [2]
>= [0 1] V1 + [7]
[0 0] [2]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 1] V1 + [1 1] V2 + [12]
[0 0] [0 0] [2]
>= [0 1] V2 + [12]
[0 0] [2]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[activate^#(n__isNat(X))] = [0 2] X + [8]
[0 0] [0]
> [0 2] X + [7]
[0 0] [0]
= [c_2(isNat^#(X))]
[isNat^#(n__plus(V1, V2))] = [2 2] V1 + [2 2] V2 + [15]
[1 1] [1 1] [9]
> [2 2] V1 + [2 2] V2 + [10]
[0 0] [0 0] [0]
= [c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))]
[isNat^#(n__s(V1))] = [2 2] V1 + [1]
[1 1] [2]
>= [2 2] V1 + [1]
[0 0] [0]
= [c_4(isNat^#(activate(V1)), activate^#(V1))]
[isNat^#(n__x(V1, V2))] = [2 2] V1 + [2 2] V2 + [11]
[1 1] [1 1] [7]
> [2 2] V1 + [2 2] V2 + [10]
[0 0] [0 0] [0]
= [c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))]
[and^#(tt(), X)] = [2 0] X + [0]
[0 0] [1]
>= [2 0] X + [0]
[0 0] [0]
= [c_8(activate^#(X))]
We return to the main proof. Consider the set of all dependency
pairs
:
{ 1: activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, 2: isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1))
, 3: isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 4: isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, 5: and^#(tt(), X) -> c_8(activate^#(X)) }
Processor 'matrix interpretation of dimension 2' induces the
complexity certificate YES(?,O(n^1)) on application of dependency
pairs {1,3,4}. These cover all (indirect) predecessors of
dependency pairs {1,3,4,5}, their number of application is equally
bounded. The dependency pairs are shifted into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) }
Weak DPs:
{ activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, and^#(tt(), X) -> c_8(activate^#(X)) }
Weak Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
DPs:
{ 1: isNat^#(n__s(V1)) ->
c_4(isNat^#(activate(V1)), activate^#(V1))
, 3: isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {1}, Uargs(c_3) = {1, 2, 3, 4}, Uargs(c_4) = {1, 2},
Uargs(c_5) = {1, 2, 3, 4}, Uargs(c_8) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[tt] = [0]
[0]
[activate](x1) = [1 0] x1 + [0]
[0 3] [0]
[s](x1) = [1 1] x1 + [1]
[0 0] [2]
[plus](x1, x2) = [1 1] x1 + [1 1] x2 + [4]
[0 0] [0 0] [2]
[0] = [0]
[0]
[x](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 0] [0 0] [1]
[and](x1, x2) = [5 0] x2 + [0]
[0 3] [0]
[isNat](x1) = [0 0] x1 + [0]
[3 0] [0]
[n__0] = [0]
[0]
[n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [4]
[0 0] [0 0] [1]
[n__isNat](x1) = [0 0] x1 + [0]
[1 0] [0]
[n__s](x1) = [1 1] x1 + [1]
[0 0] [1]
[n__x](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 0] [0 0] [1]
[activate^#](x1) = [0 1] x1 + [0]
[1 0] [0]
[isNat^#](x1) = [1 0] x1 + [0]
[0 0] [0]
[and^#](x1, x2) = [2 1] x2 + [0]
[0 0] [1]
[c_2](x1) = [1 1] x1 + [0]
[0 0] [0]
[c_3](x1, x2, x3, x4) = [1 3] x1 + [1 0] x2 + [1 0] x3 + [1
0] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[c_5](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1
0] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[c_8](x1) = [1 2] x1 + [0]
[0 0] [0]
The order satisfies the following ordering constraints:
[activate(X)] = [1 0] X + [0]
[0 3] [0]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [0]
[0]
>= [0]
[0]
= [0()]
[activate(n__plus(X1, X2))] = [1 1] X1 + [1 1] X2 + [4]
[0 0] [0 0] [3]
>= [1 1] X1 + [1 1] X2 + [4]
[0 0] [0 0] [2]
= [plus(X1, X2)]
[activate(n__isNat(X))] = [0 0] X + [0]
[3 0] [0]
>= [0 0] X + [0]
[3 0] [0]
= [isNat(X)]
[activate(n__s(X))] = [1 1] X + [1]
[0 0] [3]
>= [1 1] X + [1]
[0 0] [2]
= [s(X)]
[activate(n__x(X1, X2))] = [1 1] X1 + [1 1] X2 + [0]
[0 0] [0 0] [3]
>= [1 1] X1 + [1 1] X2 + [0]
[0 0] [0 0] [1]
= [x(X1, X2)]
[s(X)] = [1 1] X + [1]
[0 0] [2]
>= [1 1] X + [1]
[0 0] [1]
= [n__s(X)]
[plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [4]
[0 0] [0 0] [2]
>= [1 1] X1 + [1 1] X2 + [4]
[0 0] [0 0] [1]
= [n__plus(X1, X2)]
[0()] = [0]
[0]
>= [0]
[0]
= [n__0()]
[x(X1, X2)] = [1 1] X1 + [1 1] X2 + [0]
[0 0] [0 0] [1]
>= [1 1] X1 + [1 1] X2 + [0]
[0 0] [0 0] [1]
= [n__x(X1, X2)]
[and(tt(), X)] = [5 0] X + [0]
[0 3] [0]
>= [1 0] X + [0]
[0 3] [0]
= [activate(X)]
[isNat(X)] = [0 0] X + [0]
[3 0] [0]
>= [0 0] X + [0]
[1 0] [0]
= [n__isNat(X)]
[isNat(n__0())] = [0]
[0]
>= [0]
[0]
= [tt()]
[isNat(n__plus(V1, V2))] = [0 0] V1 + [0 0] V2 + [0]
[3 3] [3 3] [12]
>= [0 0] V2 + [0]
[3 0] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [0 0] V1 + [0]
[3 3] [3]
>= [0 0] V1 + [0]
[3 0] [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [0 0] V1 + [0 0] V2 + [0]
[3 3] [3 3] [0]
>= [0 0] V2 + [0]
[3 0] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[activate^#(n__isNat(X))] = [1 0] X + [0]
[0 0] [0]
>= [1 0] X + [0]
[0 0] [0]
= [c_2(isNat^#(X))]
[isNat^#(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [4]
[0 0] [0 0] [0]
> [1 1] V1 + [1 1] V2 + [3]
[0 0] [0 0] [0]
= [c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))]
[isNat^#(n__s(V1))] = [1 1] V1 + [1]
[0 0] [0]
> [1 1] V1 + [0]
[0 0] [0]
= [c_4(isNat^#(activate(V1)), activate^#(V1))]
[isNat^#(n__x(V1, V2))] = [1 1] V1 + [1 1] V2 + [0]
[0 0] [0 0] [0]
>= [1 1] V1 + [1 1] V2 + [0]
[0 0] [0 0] [0]
= [c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))]
[and^#(tt(), X)] = [2 1] X + [0]
[0 0] [1]
>= [2 1] X + [0]
[0 0] [0]
= [c_8(activate^#(X))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, and^#(tt(), X) -> c_8(activate^#(X)) }
Weak Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
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.
{ activate^#(n__isNat(X)) -> c_2(isNat^#(X))
, isNat^#(n__plus(V1, V2)) ->
c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1))
, isNat^#(n__x(V1, V2)) ->
c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))),
isNat^#(activate(V1)),
activate^#(V1),
activate^#(V2))
, and^#(tt(), X) -> c_8(activate^#(X)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(X1, X2)
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(X)
, activate(n__x(X1, X2)) -> x(X1, X2)
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, and(tt(), X) -> activate(X)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt()
, isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2)))
, isNat(n__s(V1)) -> isNat(activate(V1))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
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(n^1))