Problem Mixed HO 10 prenex

Tool Bounds

Execution Time	3.1744003e-2ms
Answer	YES(?,O(n^1))
Input	Mixed HO 10 prenex

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  not(exists()) -> forall()
     , or(exists(), P) -> exists()
     , and(exists(), P) -> exists()
     , or(P, exists()) -> exists()
     , and(P, exists()) -> exists()
     , not(forall()) -> exists()
     , or(forall(), P) -> forall()
     , and(forall(), P) -> forall()
     , or(P, forall()) -> forall()
     , and(P, forall()) -> forall()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  The problem is match-bounded by 1.
  The enriched problem is compatible with the following automaton:
  {  forall_0() -> 1
   , forall_1() -> 1
   , and_0(1, 1) -> 1
   , or_0(1, 1) -> 1
   , not_0(1) -> 1
   , exists_0() -> 1
   , exists_1() -> 1}

Hurray, we answered YES(?,O(n^1))

Tool CDI

Execution Time	0.105455875ms
Answer	YES(?,O(n^2))
Input	Mixed HO 10 prenex

stdout:

YES(?,O(n^2))
QUADRATIC upper bound on the derivational complexity

This TRS is terminating using the deltarestricted interpretation
not(delta, X0) =  + 1*X0 + 0 + 2*X0*delta + 0*delta
exists(delta) =  + 2 + 0*delta
or(delta, X1, X0) =  + 1*X0 + 1*X1 + 0 + 2*X0*delta + 3*X1*delta + 0*delta
and(delta, X1, X0) =  + 1*X0 + 1*X1 + 3 + 0*X0*delta + 1*X1*delta + 1*delta
forall(delta) =  + 2 + 0*delta
not_tau_1(delta) = delta/(1 + 2 * delta)
or_tau_1(delta) = delta/(1 + 3 * delta)
or_tau_2(delta) = delta/(1 + 2 * delta)
and_tau_1(delta) = delta/(1 + 1 * delta)
and_tau_2(delta) = delta/(1 + 0 * delta)

Time: 0.066372 seconds
Statistics:
Number of monomials: 165
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.135504ms
Answer	YES(?,O(n^1))
Input	Mixed HO 10 prenex

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  not(exists()) -> forall()
     , or(exists(), P) -> exists()
     , and(exists(), P) -> exists()
     , or(P, exists()) -> exists()
     , and(P, exists()) -> exists()
     , not(forall()) -> exists()
     , or(forall(), P) -> forall()
     , and(forall(), P) -> forall()
     , or(P, forall()) -> forall()
     , and(P, forall()) -> forall()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   forall() = [0]
   and(x1, x2) = [1] x1 + [1] x2 + [1]
   or(x1, x2) = [1] x1 + [1] x2 + [1]
   not(x1) = [1] x1 + [1]
   exists() = [0]

Hurray, we answered YES(?,O(n^1))

Tool IDA

Execution Time	0.136096ms
Answer	YES(?,O(n^1))
Input	Mixed HO 10 prenex

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  not(exists()) -> forall()
     , or(exists(), P) -> exists()
     , and(exists(), P) -> exists()
     , or(P, exists()) -> exists()
     , and(P, exists()) -> exists()
     , not(forall()) -> exists()
     , or(forall(), P) -> forall()
     , and(forall(), P) -> forall()
     , or(P, forall()) -> forall()
     , and(P, forall()) -> forall()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
  Interpretation Functions:
   forall() = [0]
   and(x1, x2) = [1] x1 + [1] x2 + [1]
   or(x1, x2) = [1] x1 + [1] x2 + [1]
   not(x1) = [1] x1 + [1]
   exists() = [0]

Hurray, we answered YES(?,O(n^1))

Tool TRI

Execution Time	9.6080065e-2ms
Answer	YES(?,O(n^1))
Input	Mixed HO 10 prenex

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  not(exists()) -> forall()
     , or(exists(), P) -> exists()
     , and(exists(), P) -> exists()
     , or(P, exists()) -> exists()
     , and(P, exists()) -> exists()
     , not(forall()) -> exists()
     , or(forall(), P) -> forall()
     , and(forall(), P) -> forall()
     , or(P, forall()) -> forall()
     , and(P, forall()) -> forall()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   forall() = [0]
   and(x1, x2) = [1] x1 + [1] x2 + [1]
   or(x1, x2) = [1] x1 + [1] x2 + [1]
   not(x1) = [1] x1 + [1]
   exists() = [0]

Hurray, we answered YES(?,O(n^1))

Tool TRI2

Execution Time	0.115441084ms
Answer	YES(?,O(n^1))
Input	Mixed HO 10 prenex

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  not(exists()) -> forall()
     , or(exists(), P) -> exists()
     , and(exists(), P) -> exists()
     , or(P, exists()) -> exists()
     , and(P, exists()) -> exists()
     , not(forall()) -> exists()
     , or(forall(), P) -> forall()
     , and(forall(), P) -> forall()
     , or(P, forall()) -> forall()
     , and(P, forall()) -> forall()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   forall() = [2]
              [0]
   and(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                 [0 0]      [0 0]      [0]
   or(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                [0 0]      [0 0]      [0]
   not(x1) = [1 0] x1 + [2]
             [0 0]      [0]
   exists() = [2]
              [0]

Hurray, we answered YES(?,O(n^1))