Problem SK90 4.01

Tool Bounds

Execution Time	6.007099e-2ms
Answer	YES(?,O(n^1))
Input	SK90 4.01

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  +(+(x, y), minus(y)) -> x
     , +(minus(x), +(x, y)) -> y
     , minux(+(x, y)) -> +(minus(y), minus(x))
     , minus(minus(x)) -> x}
  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:
  {  minus_0(1) -> 1
   , minus_0(1) -> 2
   , minus_0(1) -> 3
   , minus_1(1) -> 1
   , minus_1(1) -> 2
   , minus_1(1) -> 3
   , minus_1(2) -> 1
   , minus_1(2) -> 2
   , minus_1(2) -> 3
   , minus_1(3) -> 1
   , minus_1(3) -> 2
   , minus_1(3) -> 3
   , +_0(1, 1) -> 1
   , +_0(1, 1) -> 2
   , +_0(1, 1) -> 3
   , +_1(2, 3) -> 1
   , +_1(2, 3) -> 2
   , +_1(2, 3) -> 3
   , minux_0(1) -> 1
   , minux_0(1) -> 2
   , minux_0(1) -> 3}

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

Tool CDI

Execution Time	0.61936903ms
Answer	YES(?,O(n^2))
Input	SK90 4.01

stdout:

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

This TRS is terminating using the deltarestricted interpretation
minux(delta, X0) =  + 1*X0 + 3 + 0*X0*delta + 3*delta
+(delta, X1, X0) =  + 1*X0 + 1*X1 + 1 + 0*X0*delta + 0*X1*delta + 0*delta
minus(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 1*delta
minux_tau_1(delta) = delta/(1 + 0 * delta)
+_tau_1(delta) = delta/(1 + 0 * delta)
+_tau_2(delta) = delta/(1 + 0 * delta)
minus_tau_1(delta) = delta/(1 + 0 * delta)

Time: 0.577445 seconds
Statistics:
Number of monomials: 344
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.13389277ms
Answer	YES(?,O(n^1))
Input	SK90 4.01

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  +(+(x, y), minus(y)) -> x
     , +(minus(x), +(x, y)) -> y
     , minux(+(x, y)) -> +(minus(y), minus(x))
     , minus(minus(x)) -> x}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   minus(x1) = [1] x1 + [1]
   +(x1, x2) = [1] x1 + [1] x2 + [3]
   minux(x1) = [1] x1 + [3]

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

Tool IDA

Execution Time	0.23336911ms
Answer	YES(?,O(n^1))
Input	SK90 4.01

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  +(+(x, y), minus(y)) -> x
     , +(minus(x), +(x, y)) -> y
     , minux(+(x, y)) -> +(minus(y), minus(x))
     , minus(minus(x)) -> x}
  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:
   minus(x1) = [1] x1 + [1]
   +(x1, x2) = [1] x1 + [1] x2 + [1]
   minux(x1) = [1] x1 + [3]

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

Tool TRI

Execution Time	6.276989e-2ms
Answer	YES(?,O(n^1))
Input	SK90 4.01

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  +(+(x, y), minus(y)) -> x
     , +(minus(x), +(x, y)) -> y
     , minux(+(x, y)) -> +(minus(y), minus(x))
     , minus(minus(x)) -> x}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   minus(x1) = [1] x1 + [1]
   +(x1, x2) = [1] x1 + [1] x2 + [0]
   minux(x1) = [1] x1 + [3]

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

Tool TRI2

Execution Time	0.11964202ms
Answer	YES(?,O(n^1))
Input	SK90 4.01

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  +(+(x, y), minus(y)) -> x
     , +(minus(x), +(x, y)) -> y
     , minux(+(x, y)) -> +(minus(y), minus(x))
     , minus(minus(x)) -> x}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   minus(x1) = [1 0] x1 + [1]
               [0 1]      [0]
   +(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
               [0 1]      [0 1]      [0]
   minux(x1) = [1 0] x1 + [3]
               [0 1]      [3]

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