We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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^2))
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^2)).
Strict Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) ->
plus(x(activate(N), activate(M)), activate(N)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7] x2 + [4]
[tt] = [0]
[activate](x1) = [1] x1 + [0]
[U21](x1, x2, x3) = [7] x2 + [7] x3 + [5]
[s](x1) = [1] x1 + [0]
[plus](x1, x2) = [1] x1 + [1] x2 + [0]
[U31](x1) = [4]
[0] = [0]
[U41](x1, x2, x3) = [7] x2 + [7] x3 + [5]
[x](x1, x2) = [1] x1 + [1] x2 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[isNat](x1) = [1] x1 + [0]
[n__0] = [0]
[n__plus](x1, x2) = [1] x1 + [1] x2 + [0]
[n__isNat](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
[n__x](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7] N + [4]
> [1] N + [0]
= [activate(N)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1] X + [0]
>= [1] X + [0]
= [isNat(X)]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7] N + [7] M + [5]
> [1] N + [1] M + [0]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[plus(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__plus(X1, X2)]
[U31(tt())] = [4]
> [0]
= [0()]
[0()] = [0]
>= [0]
= [n__0()]
[U41(tt(), M, N)] = [7] N + [7] M + [5]
> [2] N + [1] M + [0]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__x(X1, X2)]
[and(tt(), X)] = [1] X + [0]
>= [1] X + [0]
= [activate(X)]
[isNat(X)] = [1] X + [0]
>= [1] X + [0]
= [n__isNat(X)]
[isNat(n__0())] = [0]
>= [0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [0]
>= [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1] V1 + [0]
>= [1] V1 + [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1] V1 + [1] V2 + [0]
>= [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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))) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) ->
plus(x(activate(N), activate(M)), activate(N)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7] x2 + [5]
[tt] = [0]
[activate](x1) = [1] x1 + [0]
[U21](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[s](x1) = [1] x1 + [0]
[plus](x1, x2) = [1] x1 + [1] x2 + [0]
[U31](x1) = [5]
[0] = [0]
[U41](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[x](x1, x2) = [1] x1 + [1] x2 + [1]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[isNat](x1) = [1] x1 + [0]
[n__0] = [0]
[n__plus](x1, x2) = [1] x1 + [1] x2 + [0]
[n__isNat](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
[n__x](x1, x2) = [1] x1 + [1] x2 + [1]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7] N + [5]
> [1] N + [0]
= [activate(N)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1] X + [0]
>= [1] X + [0]
= [isNat(X)]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7] N + [7] M + [1]
> [1] N + [1] M + [0]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[plus(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__plus(X1, X2)]
[U31(tt())] = [5]
> [0]
= [0()]
[0()] = [0]
>= [0]
= [n__0()]
[U41(tt(), M, N)] = [7] N + [7] M + [1]
>= [2] N + [1] M + [1]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__x(X1, X2)]
[and(tt(), X)] = [1] X + [0]
>= [1] X + [0]
= [activate(X)]
[isNat(X)] = [1] X + [0]
>= [1] X + [0]
= [n__isNat(X)]
[isNat(n__0())] = [0]
>= [0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [0]
>= [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1] V1 + [0]
>= [1] V1 + [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1] V1 + [1] V2 + [1]
> [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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)) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, isNat(n__x(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { isNat(n__s(V1)) -> isNat(activate(V1)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7] x2 + [5]
[tt] = [0]
[activate](x1) = [1] x1 + [0]
[U21](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[s](x1) = [1] x1 + [1]
[plus](x1, x2) = [1] x1 + [1] x2 + [0]
[U31](x1) = [4]
[0] = [0]
[U41](x1, x2, x3) = [7] x2 + [7] x3 + [4]
[x](x1, x2) = [1] x1 + [1] x2 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[isNat](x1) = [1] x1 + [0]
[n__0] = [0]
[n__plus](x1, x2) = [1] x1 + [1] x2 + [0]
[n__isNat](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [1]
[n__x](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7] N + [5]
> [1] N + [0]
= [activate(N)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1] X + [0]
>= [1] X + [0]
= [isNat(X)]
[activate(n__s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7] N + [7] M + [1]
>= [1] N + [1] M + [1]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1] X + [1]
>= [1] X + [1]
= [n__s(X)]
[plus(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__plus(X1, X2)]
[U31(tt())] = [4]
> [0]
= [0()]
[0()] = [0]
>= [0]
= [n__0()]
[U41(tt(), M, N)] = [7] N + [7] M + [4]
> [2] N + [1] M + [0]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__x(X1, X2)]
[and(tt(), X)] = [1] X + [0]
>= [1] X + [0]
= [activate(X)]
[isNat(X)] = [1] X + [0]
>= [1] X + [0]
= [n__isNat(X)]
[isNat(n__0())] = [0]
>= [0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [0]
>= [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1] V1 + [1]
> [1] V1 + [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1] V1 + [1] V2 + [0]
>= [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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))) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, 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^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ isNat(n__plus(V1, V2)) ->
and(isNat(activate(V1)), n__isNat(activate(V2))) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7] x2 + [4]
[tt] = [0]
[activate](x1) = [1] x1 + [0]
[U21](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[s](x1) = [1] x1 + [0]
[plus](x1, x2) = [1] x1 + [1] x2 + [1]
[U31](x1) = [4]
[0] = [0]
[U41](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[x](x1, x2) = [1] x1 + [1] x2 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[isNat](x1) = [1] x1 + [0]
[n__0] = [0]
[n__plus](x1, x2) = [1] x1 + [1] x2 + [1]
[n__isNat](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
[n__x](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7] N + [4]
> [1] N + [0]
= [activate(N)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1] X + [0]
>= [1] X + [0]
= [isNat(X)]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7] N + [7] M + [1]
>= [1] N + [1] M + [1]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[plus(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__plus(X1, X2)]
[U31(tt())] = [4]
> [0]
= [0()]
[0()] = [0]
>= [0]
= [n__0()]
[U41(tt(), M, N)] = [7] N + [7] M + [1]
>= [2] N + [1] M + [1]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__x(X1, X2)]
[and(tt(), X)] = [1] X + [0]
>= [1] X + [0]
= [activate(X)]
[isNat(X)] = [1] X + [0]
>= [1] X + [0]
= [n__isNat(X)]
[isNat(n__0())] = [0]
>= [0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [1]
> [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1] V1 + [0]
>= [1] V1 + [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1] V1 + [1] V2 + [0]
>= [1] V1 + [1] V2 + [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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() }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, 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^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { and(tt(), X) -> activate(X) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7] x2 + [5]
[tt] = [1]
[activate](x1) = [1] x1 + [0]
[U21](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[s](x1) = [1] x1 + [0]
[plus](x1, x2) = [1] x1 + [1] x2 + [1]
[U31](x1) = [5]
[0] = [0]
[U41](x1, x2, x3) = [7] x2 + [7] x3 + [3]
[x](x1, x2) = [1] x1 + [1] x2 + [1]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[isNat](x1) = [1] x1 + [1]
[n__0] = [0]
[n__plus](x1, x2) = [1] x1 + [1] x2 + [1]
[n__isNat](x1) = [1] x1 + [1]
[n__s](x1) = [1] x1 + [0]
[n__x](x1, x2) = [1] x1 + [1] x2 + [1]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7] N + [5]
> [1] N + [0]
= [activate(N)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1] X + [1]
>= [1] X + [1]
= [isNat(X)]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7] N + [7] M + [1]
>= [1] N + [1] M + [1]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[plus(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__plus(X1, X2)]
[U31(tt())] = [5]
> [0]
= [0()]
[0()] = [0]
>= [0]
= [n__0()]
[U41(tt(), M, N)] = [7] N + [7] M + [3]
> [2] N + [1] M + [2]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__x(X1, X2)]
[and(tt(), X)] = [1] X + [1]
> [1] X + [0]
= [activate(X)]
[isNat(X)] = [1] X + [1]
>= [1] X + [1]
= [n__isNat(X)]
[isNat(n__0())] = [1]
>= [1]
= [tt()]
[isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [2]
>= [1] V1 + [1] V2 + [2]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1] V1 + [1]
>= [1] V1 + [1]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1] V1 + [1] V2 + [2]
>= [1] V1 + [1] V2 + [2]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, isNat(X) -> n__isNat(X)
, isNat(n__0()) -> tt() }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, and(tt(), X) -> activate(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))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { isNat(n__0()) -> tt() }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7] x2 + [5]
[tt] = [0]
[activate](x1) = [1] x1 + [0]
[U21](x1, x2, x3) = [7] x2 + [7] x3 + [1]
[s](x1) = [1] x1 + [0]
[plus](x1, x2) = [1] x1 + [1] x2 + [1]
[U31](x1) = [5]
[0] = [0]
[U41](x1, x2, x3) = [7] x2 + [7] x3 + [3]
[x](x1, x2) = [1] x1 + [1] x2 + [1]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[isNat](x1) = [1] x1 + [1]
[n__0] = [0]
[n__plus](x1, x2) = [1] x1 + [1] x2 + [1]
[n__isNat](x1) = [1] x1 + [1]
[n__s](x1) = [1] x1 + [0]
[n__x](x1, x2) = [1] x1 + [1] x2 + [1]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7] N + [5]
> [1] N + [0]
= [activate(N)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1] X + [1]
>= [1] X + [1]
= [isNat(X)]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7] N + [7] M + [1]
>= [1] N + [1] M + [1]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[plus(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__plus(X1, X2)]
[U31(tt())] = [5]
> [0]
= [0()]
[0()] = [0]
>= [0]
= [n__0()]
[U41(tt(), M, N)] = [7] N + [7] M + [3]
> [2] N + [1] M + [2]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__x(X1, X2)]
[and(tt(), X)] = [1] X + [0]
>= [1] X + [0]
= [activate(X)]
[isNat(X)] = [1] X + [1]
>= [1] X + [1]
= [n__isNat(X)]
[isNat(n__0())] = [1]
> [0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [2]
>= [1] V1 + [1] V2 + [2]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1] V1 + [1]
>= [1] V1 + [1]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1] V1 + [1] V2 + [2]
>= [1] V1 + [1] V2 + [2]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, 0() -> n__0()
, x(X1, X2) -> n__x(X1, X2)
, isNat(X) -> n__isNat(X) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
, and(tt(), X) -> activate(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^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, 0() -> n__0()
, isNat(X) -> n__isNat(X) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7 7] x2 + [7]
[7 7] [4]
[tt] = [2]
[0]
[activate](x1) = [1 1] x1 + [2]
[0 1] [0]
[U21](x1, x2, x3) = [4 0] x1 + [7 7] x2 + [7 7] x3 + [7]
[0 0] [7 7] [7 7] [7]
[s](x1) = [1 1] x1 + [6]
[0 1] [0]
[plus](x1, x2) = [1 1] x1 + [1 1] x2 + [2]
[0 1] [0 1] [3]
[U31](x1) = [5]
[4]
[0] = [2]
[0]
[U41](x1, x2, x3) = [7 0] x1 + [7 7] x2 + [7 7] x3 + [1]
[0 0] [7 7] [7 7] [5]
[x](x1, x2) = [1 1] x1 + [1 1] x2 + [4]
[0 1] [0 1] [2]
[and](x1, x2) = [1 0] x1 + [1 1] x2 + [1]
[0 0] [0 1] [0]
[isNat](x1) = [1 1] x1 + [1]
[0 1] [0]
[n__0] = [1]
[0]
[n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [2]
[0 1] [0 1] [3]
[n__isNat](x1) = [1 0] x1 + [0]
[0 1] [0]
[n__s](x1) = [1 1] x1 + [6]
[0 1] [0]
[n__x](x1, x2) = [1 1] x1 + [1 1] x2 + [4]
[0 1] [0 1] [2]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7 7] N + [7]
[7 7] [4]
> [1 1] N + [2]
[0 1] [0]
= [activate(N)]
[activate(X)] = [1 1] X + [2]
[0 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [3]
[0]
> [2]
[0]
= [0()]
[activate(n__plus(X1, X2))] = [1 2] X1 + [1 2] X2 + [7]
[0 1] [0 1] [3]
> [1 2] X1 + [1 2] X2 + [6]
[0 1] [0 1] [3]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1 1] X + [2]
[0 1] [0]
> [1 1] X + [1]
[0 1] [0]
= [isNat(X)]
[activate(n__s(X))] = [1 2] X + [8]
[0 1] [0]
>= [1 2] X + [8]
[0 1] [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1 2] X1 + [1 2] X2 + [8]
[0 1] [0 1] [2]
>= [1 2] X1 + [1 2] X2 + [8]
[0 1] [0 1] [2]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7 7] N + [7 7] M + [15]
[7 7] [7 7] [7]
>= [1 3] N + [1 3] M + [15]
[0 1] [0 1] [3]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1 1] X + [6]
[0 1] [0]
>= [1 1] X + [6]
[0 1] [0]
= [n__s(X)]
[plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [2]
[0 1] [0 1] [3]
>= [1 1] X1 + [1 1] X2 + [2]
[0 1] [0 1] [3]
= [n__plus(X1, X2)]
[U31(tt())] = [5]
[4]
> [2]
[0]
= [0()]
[0()] = [2]
[0]
> [1]
[0]
= [n__0()]
[U41(tt(), M, N)] = [7 7] N + [7 7] M + [15]
[7 7] [7 7] [5]
> [2 5] N + [1 3] M + [14]
[0 2] [0 1] [5]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1 1] X1 + [1 1] X2 + [4]
[0 1] [0 1] [2]
>= [1 1] X1 + [1 1] X2 + [4]
[0 1] [0 1] [2]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 1] X + [3]
[0 1] [0]
> [1 1] X + [2]
[0 1] [0]
= [activate(X)]
[isNat(X)] = [1 1] X + [1]
[0 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [n__isNat(X)]
[isNat(n__0())] = [2]
[0]
>= [2]
[0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 2] V1 + [1 2] V2 + [6]
[0 1] [0 1] [3]
>= [1 2] V1 + [1 2] V2 + [6]
[0 0] [0 1] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 2] V1 + [7]
[0 1] [0]
> [1 2] V1 + [3]
[0 1] [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 2] V1 + [1 2] V2 + [7]
[0 1] [0 1] [2]
> [1 2] V1 + [1 2] V2 + [6]
[0 0] [0 1] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, x(X1, X2) -> n__x(X1, X2) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(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^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs: { activate(n__s(X)) -> s(activate(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7 7] x2 + [7]
[7 7] [4]
[tt] = [1]
[0]
[activate](x1) = [1 1] x1 + [0]
[0 1] [0]
[U21](x1, x2, x3) = [3 0] x1 + [7 7] x2 + [7 7] x3 + [7]
[0 0] [7 7] [7 7] [4]
[s](x1) = [1 1] x1 + [7]
[0 1] [1]
[plus](x1, x2) = [1 2] x1 + [1 1] x2 + [2]
[0 1] [0 1] [1]
[U31](x1) = [7]
[4]
[0] = [4]
[1]
[U41](x1, x2, x3) = [7 7] x2 + [7 7] x3 + [3]
[7 7] [7 7] [5]
[x](x1, x2) = [1 2] x1 + [1 4] x2 + [1]
[0 1] [0 1] [0]
[and](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 0] [0 1] [0]
[isNat](x1) = [1 5] x1 + [0]
[0 1] [0]
[n__0] = [3]
[1]
[n__plus](x1, x2) = [1 2] x1 + [1 1] x2 + [2]
[0 1] [0 1] [1]
[n__isNat](x1) = [1 4] x1 + [0]
[0 1] [0]
[n__s](x1) = [1 1] x1 + [7]
[0 1] [1]
[n__x](x1, x2) = [1 2] x1 + [1 4] x2 + [1]
[0 1] [0 1] [0]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7 7] N + [7]
[7 7] [4]
> [1 1] N + [0]
[0 1] [0]
= [activate(N)]
[activate(X)] = [1 1] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [4]
[1]
>= [4]
[1]
= [0()]
[activate(n__plus(X1, X2))] = [1 3] X1 + [1 2] X2 + [3]
[0 1] [0 1] [1]
> [1 3] X1 + [1 2] X2 + [2]
[0 1] [0 1] [1]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1 5] X + [0]
[0 1] [0]
>= [1 5] X + [0]
[0 1] [0]
= [isNat(X)]
[activate(n__s(X))] = [1 2] X + [8]
[0 1] [1]
> [1 2] X + [7]
[0 1] [1]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1 3] X1 + [1 5] X2 + [1]
[0 1] [0 1] [0]
>= [1 3] X1 + [1 5] X2 + [1]
[0 1] [0 1] [0]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7 7] N + [7 7] M + [10]
[7 7] [7 7] [4]
>= [1 4] N + [1 3] M + [10]
[0 1] [0 1] [2]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1 1] X + [7]
[0 1] [1]
>= [1 1] X + [7]
[0 1] [1]
= [n__s(X)]
[plus(X1, X2)] = [1 2] X1 + [1 1] X2 + [2]
[0 1] [0 1] [1]
>= [1 2] X1 + [1 1] X2 + [2]
[0 1] [0 1] [1]
= [n__plus(X1, X2)]
[U31(tt())] = [7]
[4]
> [4]
[1]
= [0()]
[0()] = [4]
[1]
> [3]
[1]
= [n__0()]
[U41(tt(), M, N)] = [7 7] N + [7 7] M + [3]
[7 7] [7 7] [5]
>= [2 7] N + [1 7] M + [3]
[0 2] [0 1] [1]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1 2] X1 + [1 4] X2 + [1]
[0 1] [0 1] [0]
>= [1 2] X1 + [1 4] X2 + [1]
[0 1] [0 1] [0]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 1] X + [1]
[0 1] [0]
> [1 1] X + [0]
[0 1] [0]
= [activate(X)]
[isNat(X)] = [1 5] X + [0]
[0 1] [0]
>= [1 4] X + [0]
[0 1] [0]
= [n__isNat(X)]
[isNat(n__0())] = [8]
[1]
> [1]
[0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 7] V1 + [1 6] V2 + [7]
[0 1] [0 1] [1]
> [1 7] V1 + [1 6] V2 + [0]
[0 0] [0 1] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 6] V1 + [12]
[0 1] [1]
> [1 6] V1 + [0]
[0 1] [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 7] V1 + [1 9] V2 + [1]
[0 1] [0 1] [0]
> [1 7] V1 + [1 6] V2 + [0]
[0 0] [0 1] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
, s(X) -> n__s(X)
, plus(X1, X2) -> n__plus(X1, X2)
, x(X1, X2) -> n__x(X1, X2) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(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^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
, s(X) -> n__s(X) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7 7] x2 + [6]
[7 7] [4]
[tt] = [2]
[3]
[activate](x1) = [1 1] x1 + [2]
[0 1] [0]
[U21](x1, x2, x3) = [4 0] x1 + [7 7] x2 + [7 7] x3 + [5]
[4 0] [7 7] [7 7] [0]
[s](x1) = [1 1] x1 + [4]
[0 1] [3]
[plus](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 1] [0 1] [5]
[U31](x1) = [4]
[5]
[0] = [1]
[5]
[U41](x1, x2, x3) = [4 0] x1 + [7 7] x2 + [7 7] x3 + [7]
[4 0] [7 7] [7 7] [1]
[x](x1, x2) = [1 1] x1 + [1 1] x2 + [5]
[0 1] [0 1] [3]
[and](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 0] [0 1] [2]
[isNat](x1) = [1 2] x1 + [3]
[0 1] [5]
[n__0] = [0]
[5]
[n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 1] [0 1] [5]
[n__isNat](x1) = [1 1] x1 + [0]
[0 1] [5]
[n__s](x1) = [1 1] x1 + [2]
[0 1] [3]
[n__x](x1, x2) = [1 1] x1 + [1 1] x2 + [5]
[0 1] [0 1] [3]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7 7] N + [6]
[7 7] [4]
> [1 1] N + [2]
[0 1] [0]
= [activate(N)]
[activate(X)] = [1 1] X + [2]
[0 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [7]
[5]
> [1]
[5]
= [0()]
[activate(n__plus(X1, X2))] = [1 2] X1 + [1 2] X2 + [7]
[0 1] [0 1] [5]
> [1 2] X1 + [1 2] X2 + [4]
[0 1] [0 1] [5]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1 2] X + [7]
[0 1] [5]
> [1 2] X + [3]
[0 1] [5]
= [isNat(X)]
[activate(n__s(X))] = [1 2] X + [7]
[0 1] [3]
> [1 2] X + [6]
[0 1] [3]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1 2] X1 + [1 2] X2 + [10]
[0 1] [0 1] [3]
> [1 2] X1 + [1 2] X2 + [9]
[0 1] [0 1] [3]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7 7] N + [7 7] M + [13]
[7 7] [7 7] [8]
>= [1 3] N + [1 3] M + [13]
[0 1] [0 1] [8]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1 1] X + [4]
[0 1] [3]
> [1 1] X + [2]
[0 1] [3]
= [n__s(X)]
[plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [5]
>= [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [5]
= [n__plus(X1, X2)]
[U31(tt())] = [4]
[5]
> [1]
[5]
= [0()]
[0()] = [1]
[5]
> [0]
[5]
= [n__0()]
[U41(tt(), M, N)] = [7 7] N + [7 7] M + [15]
[7 7] [7 7] [9]
> [2 5] N + [1 3] M + [14]
[0 2] [0 1] [8]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1 1] X1 + [1 1] X2 + [5]
[0 1] [0 1] [3]
>= [1 1] X1 + [1 1] X2 + [5]
[0 1] [0 1] [3]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 1] X + [2]
[0 1] [2]
>= [1 1] X + [2]
[0 1] [0]
= [activate(X)]
[isNat(X)] = [1 2] X + [3]
[0 1] [5]
> [1 1] X + [0]
[0 1] [5]
= [n__isNat(X)]
[isNat(n__0())] = [13]
[10]
> [2]
[3]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 3] V1 + [1 3] V2 + [13]
[0 1] [0 1] [10]
> [1 3] V1 + [1 3] V2 + [12]
[0 0] [0 1] [7]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 3] V1 + [11]
[0 1] [8]
> [1 3] V1 + [5]
[0 1] [5]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 3] V1 + [1 3] V2 + [14]
[0 1] [0 1] [8]
> [1 3] V1 + [1 3] V2 + [12]
[0 0] [0 1] [7]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ plus(X1, X2) -> n__plus(X1, X2)
, x(X1, X2) -> n__x(X1, X2) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
, U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
, s(X) -> n__s(X)
, U31(tt()) -> 0()
, 0() -> n__0()
, U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(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^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs: { plus(X1, X2) -> n__plus(X1, X2) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7 7] x2 + [6]
[7 7] [4]
[tt] = [1]
[0]
[activate](x1) = [1 1] x1 + [1]
[0 1] [0]
[U21](x1, x2, x3) = [4 0] x1 + [7 7] x2 + [7 7] x3 + [5]
[0 0] [7 7] [7 7] [6]
[s](x1) = [1 1] x1 + [1]
[0 1] [1]
[plus](x1, x2) = [1 1] x1 + [1 1] x2 + [1]
[0 1] [0 1] [5]
[U31](x1) = [7]
[4]
[0] = [1]
[0]
[U41](x1, x2, x3) = [7 7] x2 + [7 7] x3 + [7]
[7 7] [7 7] [7]
[x](x1, x2) = [1 1] x1 + [1 1] x2 + [1]
[0 1] [0 1] [2]
[and](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 0] [0 1] [1]
[isNat](x1) = [1 2] x1 + [0]
[0 1] [0]
[n__0] = [1]
[0]
[n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 1] [0 1] [5]
[n__isNat](x1) = [1 1] x1 + [0]
[0 1] [0]
[n__s](x1) = [1 1] x1 + [1]
[0 1] [1]
[n__x](x1, x2) = [1 1] x1 + [1 1] x2 + [1]
[0 1] [0 1] [2]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7 7] N + [6]
[7 7] [4]
> [1 1] N + [1]
[0 1] [0]
= [activate(N)]
[activate(X)] = [1 1] X + [1]
[0 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [2]
[0]
> [1]
[0]
= [0()]
[activate(n__plus(X1, X2))] = [1 2] X1 + [1 2] X2 + [6]
[0 1] [0 1] [5]
> [1 2] X1 + [1 2] X2 + [3]
[0 1] [0 1] [5]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1 2] X + [1]
[0 1] [0]
> [1 2] X + [0]
[0 1] [0]
= [isNat(X)]
[activate(n__s(X))] = [1 2] X + [3]
[0 1] [1]
> [1 2] X + [2]
[0 1] [1]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1 2] X1 + [1 2] X2 + [4]
[0 1] [0 1] [2]
> [1 2] X1 + [1 2] X2 + [3]
[0 1] [0 1] [2]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7 7] N + [7 7] M + [9]
[7 7] [7 7] [6]
>= [1 3] N + [1 3] M + [9]
[0 1] [0 1] [6]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1 1] X + [1]
[0 1] [1]
>= [1 1] X + [1]
[0 1] [1]
= [n__s(X)]
[plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [1]
[0 1] [0 1] [5]
> [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [5]
= [n__plus(X1, X2)]
[U31(tt())] = [7]
[4]
> [1]
[0]
= [0()]
[0()] = [1]
[0]
>= [1]
[0]
= [n__0()]
[U41(tt(), M, N)] = [7 7] N + [7 7] M + [7]
[7 7] [7 7] [7]
>= [2 5] N + [1 3] M + [7]
[0 2] [0 1] [7]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1 1] X1 + [1 1] X2 + [1]
[0 1] [0 1] [2]
>= [1 1] X1 + [1 1] X2 + [1]
[0 1] [0 1] [2]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 1] X + [1]
[0 1] [1]
>= [1 1] X + [1]
[0 1] [0]
= [activate(X)]
[isNat(X)] = [1 2] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [n__isNat(X)]
[isNat(n__0())] = [1]
[0]
>= [1]
[0]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 3] V1 + [1 3] V2 + [10]
[0 1] [0 1] [5]
> [1 3] V1 + [1 3] V2 + [2]
[0 0] [0 1] [1]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 3] V1 + [3]
[0 1] [1]
> [1 3] V1 + [1]
[0 1] [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 3] V1 + [1 3] V2 + [5]
[0 1] [0 1] [2]
> [1 3] V1 + [1 3] V2 + [2]
[0 0] [0 1] [1]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs: { x(X1, X2) -> n__x(X1, X2) }
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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))
, 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^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs: { x(X1, X2) -> n__x(X1, X2) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(x) = {1, 2},
Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[U11](x1, x2) = [7 7] x2 + [7]
[7 7] [4]
[tt] = [4]
[1]
[activate](x1) = [1 1] x1 + [0]
[0 1] [0]
[U21](x1, x2, x3) = [2 1] x1 + [7 7] x2 + [7 7] x3 + [3]
[0 0] [7 7] [7 7] [5]
[s](x1) = [1 3] x1 + [0]
[0 1] [0]
[plus](x1, x2) = [1 1] x1 + [1 3] x2 + [4]
[0 1] [0 1] [2]
[U31](x1) = [7]
[4]
[0] = [0]
[4]
[U41](x1, x2, x3) = [2 0] x1 + [7 7] x2 + [7 7] x3 + [7]
[0 0] [7 7] [7 7] [7]
[x](x1, x2) = [1 1] x1 + [1 4] x2 + [7]
[0 1] [0 1] [4]
[and](x1, x2) = [1 0] x1 + [1 2] x2 + [5]
[0 0] [0 1] [0]
[isNat](x1) = [1 2] x1 + [1]
[0 1] [0]
[n__0] = [0]
[4]
[n__plus](x1, x2) = [1 1] x1 + [1 3] x2 + [4]
[0 1] [0 1] [2]
[n__isNat](x1) = [1 2] x1 + [1]
[0 1] [0]
[n__s](x1) = [1 3] x1 + [0]
[0 1] [0]
[n__x](x1, x2) = [1 1] x1 + [1 4] x2 + [6]
[0 1] [0 1] [4]
The order satisfies the following ordering constraints:
[U11(tt(), N)] = [7 7] N + [7]
[7 7] [4]
> [1 1] N + [0]
[0 1] [0]
= [activate(N)]
[activate(X)] = [1 1] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [4]
[4]
> [0]
[4]
= [0()]
[activate(n__plus(X1, X2))] = [1 2] X1 + [1 4] X2 + [6]
[0 1] [0 1] [2]
> [1 2] X1 + [1 4] X2 + [4]
[0 1] [0 1] [2]
= [plus(activate(X1), activate(X2))]
[activate(n__isNat(X))] = [1 3] X + [1]
[0 1] [0]
>= [1 2] X + [1]
[0 1] [0]
= [isNat(X)]
[activate(n__s(X))] = [1 4] X + [0]
[0 1] [0]
>= [1 4] X + [0]
[0 1] [0]
= [s(activate(X))]
[activate(n__x(X1, X2))] = [1 2] X1 + [1 5] X2 + [10]
[0 1] [0 1] [4]
> [1 2] X1 + [1 5] X2 + [7]
[0 1] [0 1] [4]
= [x(activate(X1), activate(X2))]
[U21(tt(), M, N)] = [7 7] N + [7 7] M + [12]
[7 7] [7 7] [5]
> [1 5] N + [1 7] M + [10]
[0 1] [0 1] [2]
= [s(plus(activate(N), activate(M)))]
[s(X)] = [1 3] X + [0]
[0 1] [0]
>= [1 3] X + [0]
[0 1] [0]
= [n__s(X)]
[plus(X1, X2)] = [1 1] X1 + [1 3] X2 + [4]
[0 1] [0 1] [2]
>= [1 1] X1 + [1 3] X2 + [4]
[0 1] [0 1] [2]
= [n__plus(X1, X2)]
[U31(tt())] = [7]
[4]
> [0]
[4]
= [0()]
[0()] = [0]
[4]
>= [0]
[4]
= [n__0()]
[U41(tt(), M, N)] = [7 7] N + [7 7] M + [15]
[7 7] [7 7] [7]
>= [2 7] N + [1 6] M + [15]
[0 2] [0 1] [6]
= [plus(x(activate(N), activate(M)), activate(N))]
[x(X1, X2)] = [1 1] X1 + [1 4] X2 + [7]
[0 1] [0 1] [4]
> [1 1] X1 + [1 4] X2 + [6]
[0 1] [0 1] [4]
= [n__x(X1, X2)]
[and(tt(), X)] = [1 2] X + [9]
[0 1] [0]
> [1 1] X + [0]
[0 1] [0]
= [activate(X)]
[isNat(X)] = [1 2] X + [1]
[0 1] [0]
>= [1 2] X + [1]
[0 1] [0]
= [n__isNat(X)]
[isNat(n__0())] = [9]
[4]
> [4]
[1]
= [tt()]
[isNat(n__plus(V1, V2))] = [1 3] V1 + [1 5] V2 + [9]
[0 1] [0 1] [2]
> [1 3] V1 + [1 5] V2 + [7]
[0 0] [0 1] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
[isNat(n__s(V1))] = [1 5] V1 + [1]
[0 1] [0]
>= [1 3] V1 + [1]
[0 1] [0]
= [isNat(activate(V1))]
[isNat(n__x(V1, V2))] = [1 3] V1 + [1 6] V2 + [15]
[0 1] [0 1] [4]
> [1 3] V1 + [1 5] V2 + [7]
[0 0] [0 1] [0]
= [and(isNat(activate(V1)), n__isNat(activate(V2)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ U11(tt(), N) -> activate(N)
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
, activate(n__isNat(X)) -> isNat(X)
, activate(n__s(X)) -> s(activate(X))
, activate(n__x(X1, X2)) -> x(activate(X1), activate(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(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))