I simplified the original problem up to this point
((P∧¬R)∨(¬Q∨R))∧((Q∧¬R)∨(¬P∨R))
, and I got stuck here. What would be the next step? Thanks for the help!!
I am solving it with you.
Hint-1: ((P∧Q)∨R) = (PVR) ∧ (QVR)
Hint-2: P ∧ True = P
Hint-3: P V True = True
Answer
It would be true in the end. Check it once.
Next step would be
= [(P V (~Q V R)) ^ ( ~R V (~Q V R))]
^[(Q V (~P V R)) ^ ( ~R V (~P V R))]
= (P V ~Q V R) ^ ( ~p V Q V R)
= R V ( (P V ~Q) ^ ( Q V ~P))
= R v (( Q -> P ) ^ ( P -> Q))
= R V (P <-> Q)
whenever R is True it is True.
Else
P Q P<->Q
------------------
F F T
F T F
T F F
T T T
So it conforms to the truth table. Shown above by trincot.
The following expressions are equivalent:
(p ⇒ r) ⇔ (q ⇒ r)
(¬p v r) ⇔ (¬q v r)
(¬p ⇔ ¬q) v r
(p ⇔ q) v r
(p ∧ q) v (¬p ∧ ¬q) v r
Truth table:
p q r result
---------------
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
Related
I'm using nssm.exe in my scripts to manage the windows services. But in PowerShell, the command output is coming with spaces after every alphabet.
PS> $nssm = (Get-Command D:\nssm.exe)
PS> & $nssm
nssm.exe : N S S M : T h e n o n - s u c k i n g s e r v i c e m a n a g e r
At line:1 char:1
+ & $nssm
+ ~~~~~~~
+ CategoryInfo : NotSpecified: (N S S M : T h... m a n a g e r :String) [], RemoteException
+ FullyQualifiedErrorId : NativeCommandError
V e r s i o n 2 . 2 4 3 2 - b i t , 2 0 1 4 - 0 8 - 3 1
U s a g e : n s s m < o p t i o n > [ < a r g s > . . . ]
T o s h o w s e r v i c e i n s t a l l a t i o n G U I :
n s s m i n s t a l l [ < s e r v i c e n a m e > ]
T o i n s t a l l a s e r v i c e w i t h o u t c o n f i r m a t i o n :
n s s m i n s t a l l < s e r v i c e n a m e > < a p p > [ < a r g s > . . . ]
T o s h o w s e r v i c e e d i t i n g G U I :
n s s m e d i t < s e r v i c e n a m e >
How to get the output without such wide-format spaces among alphabets?
I am learning the basic part right now. And I found that the mult function contains plus as below:
Fixpoint mult (n m : nat) : nat :=
match n with
| O ⇒ O
| S n' ⇒ plus m (mult n' m)
end.
Example test_mult1: (mult 3 3) = 9.
Proof. simpl. reflexivity. Qed.
i unfold the mult and it showed
1 subgoal
______________________________________(1/1)
3 + (3 + (3 + 0)) = 9
I know that BC plus m and m is 3 so there are 3 times of plus. But I am curious where the 0 comes from. if this 3 means m, then where is n?
Thank you!
First, you should remember that 3 is a notation for S (S (S O))
When you ask for the computation of mult 3 3, only one step of mult, you get this:
mult 3 3 = mult (S 2) 3 = 3 + mult 2 3
Now if you compute mult 2 3 you get :
mult 2 3 = mult (S 1) 3 = 3 + mult 1 3
and then:
mult 1 3 = mult (S O) 3 = 3 + mult 0 3
and tnen:
mult 0 3 = 0
So the 0 that appears when computing mult comes from mult 0 3
My requirement is to find the location of a particular string in a line from notepad file but while reading it in powershell additional space get added that's why not able to find the location of a particular string. How I can find the location is this case??
I am using this code for achieving this
$ParamsPathForData = ($dir + "\TimeStats\TimeStats_1slot\29_12_2015_07TimeStats1.txt")
$data = Get-Content $ParamsPathForData
write-host $data.count total lines read from file
foreach ($line in $data)
{
$l =$line.IndexOf("12/29/2015")
write-host $l
}
I am reading this line from notepad ->
TimeStats 29 12/29/2015 7:13:42 AM +00:00 Debug PREPROCESS: SlotNo:
325-00313, Ip Address: 10.2.200.15, Duplicate Message: False,
Player-Card-No: , MessageId: 883250003130047966, MessageName:
GameIdInfo, Thread Init Delay: 14, Time To Parse: 155, Time To Exec
Main Workflow: 424, Time To Construct & send Response: 22, Total
Response Time: 615
But while exceuting it in powershell i am getting this with additinal spaces ->
T i m e S t a t s 2 9 1 2 / 2 9 / 2 0 1 5 7 : 1 3 : 4 2 A M
+ 0 0 : 0 0 D e b u g P R E P R O C E S S : S l o t N o : 3 2 5 - 0 0 3 1 3 , I p A d d r e s s : 1 0 . 2 . 2 0 0 . 1 5 , D
u p l i c a t e M e s s a g e : F a l s e , P l a y e r
- C a r d - N o : , M e s s a g e I d : 8 8 3 2 5 0 0 0 3 1 3 0 0 4 7 9 6 6 , M e s s a g e N a m e : G a m e I d I n f o , T h
r e a d I n i t D e l a y : 1 4 , T i m e T o P a r s e :
1 5 5 , T i m e T o E x e c M a i n W o r k f l o w : 4 2
4 , T i m e T o C o n s t r u c t & s e n d R e s p o n s
e : 2 2 , T o t a l R e s p o n s e T i m e : 6 1 5
Anybody please help me???
Change the the encoding to Unicode...
$data = Get-Content $ParamsPathForData -Encoding Unicode
var int a0 := 0x67452301 //A
var int b0 := 0xefcdab89 //B
var int c0 := 0x98badcfe //C
var int d0 := 0x10325476 //D
for each 512-bit chunk of message
break chunk into sixteen 32-bit words M[j], 0 ≤ j ≤ 15
//Initialize hash value for this chunk:
var int A := a0
var int B := b0
var int C := c0
var int D := d0
//Main loop:
for i from 0 to 63
if 0 ≤ i ≤ 15 then
F := (B and C) or ((not B) and D)
g := i
else if 16 ≤ i ≤ 31
F := (D and B) or ((not D) and C)
g := (5×i + 1) mod 16
else if 32 ≤ i ≤ 47
F := B xor C xor D
g := (3×i + 5) mod 16
else if 48 ≤ i ≤ 63
F := C xor (B or (not D))
g := (7×i) mod 16
dTemp := D
D := C
C := B
B := B + leftrotate((A + F + K[i] + M[g]), s[i])
A := dTemp
end for
//Add this chunk's hash to result so far:
a0 := a0 + A
b0 := b0 + B
c0 := c0 + C
d0 := d0 + D
end for
This is taken from wikipedia, specifically from here, to see the full code.
I fail to understand what for example (B and C) produces, as B and C are hexes. (Big endian)
B and C is the bitwise (32bit) boolean AND operation. So for example the int 15 (0xf) and 17 (0x11) result in 1 (0x1).
The representation of the numbers (hex or decimal) on output/input has nothing to do with the operations on them. The actual result are 4 32bit integers which get concatenated together - typically printed as as a single large hex string.
I'm working with some tools, and the only way it can determine if a particular transaction is successful is if it passes various checks. However, it is limited in the way that it can only do one check at a time, and it must be sequential. Everything must be computed from left to right.
For example,
A || C && D
It will be computed with A || C first, and then the result will be AND'ed with D.
It gets tougher with parenthesis. I am unable to compute an expression like this, since B || C would need to be compututed first. I cannot work with any order of operations;
A && ( B || C)
I think I've worked this down to this sequential boolean expression,
C || B && A
Where C || B is computed first, then that result is AND'd with A
Can all boolean expressions be successfully worked into a sequential boolean expression? (Like the example I have)
The answer is no:
Consider A || B && C || D which has the truth table:
A | B | C | D |
0 | 0 | 0 | 0 | 0
0 | 0 | 0 | 1 | 0
0 | 0 | 1 | 0 | 0
0 | 0 | 1 | 1 | 0
0 | 1 | 0 | 0 | 0
0 | 1 | 0 | 1 | 1
0 | 1 | 1 | 0 | 1
0 | 1 | 1 | 1 | 1
1 | 0 | 0 | 0 | 0
1 | 0 | 0 | 1 | 1
1 | 0 | 1 | 0 | 1
1 | 0 | 1 | 1 | 1
1 | 1 | 0 | 0 | 0
1 | 1 | 0 | 1 | 1
1 | 1 | 1 | 0 | 1
1 | 1 | 1 | 1 | 1
If it were possible to evaluate sequentially there would have to be a last expression which would be one of two cases:
Case 1:
X || Y such that Y is one of A,B,C,D and X is any sequential boolean expression.
Now, since there is no variable in A,B,C,D where the entire expression is true whenever that variable is true, none of:
X || A
X || B
X || C
X || D
can possibly be the last operation in the expression (for any X).
Case 2:
X && Y: such that Y is one of A,B,C,D and X is any sequential boolean expression.
Now, since there is no variable in A,B,C,D where the entire expression is false whenever that variable is false, none of:
X && A
X && B
X && C
X && D
can possibly be the last operation in the expression (for any X).
Therefore you cannot write (A || B) && (C || D) in this way.
The reason you are able to do this for some expressions, like: A && ( B || C) becoming C || B && A is because that expression can be built recursively out of expressions which have one of the two properties above:
IE.
The truth table for A && ( B || C) is:
A | B | C |
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0
1 | 0 | 0 | 0
1 | 0 | 1 | 1
1 | 1 | 0 | 1
1 | 1 | 1 | 1
Which we can quickly see has the property that it is false whenever A is 0. So Our expression Could be X && A.
Then we take A out of the truth table and look at only the rows where A is 1 is the original:
B | C
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 1
Which has the property that it is True whenever B is 1 (or C, we can pick here). So we can write the expression as
X || B and the entire expression becomes X || B && A
Then we reduce the table again to the portion where B was 0 and we get:
C
0 | 0
1 | 1
X is just C. So the final expression is C || B && A
This is a problem of rewriting an expression so that no parentheses occur on the right. Logical AND (∧) and OR (∨) are both commutative:
A ∧ B = B ∧ A
A ∨ B = B ∨ A
So you can rewrite any expression of the form “X a (Y)” as “(Y) a X”:
A ∧ (B ∧ C) = (A ∧ B) ∧ C
A ∧ (B ∨ C) = (B ∨ C) ∧ A
A ∨ (B ∧ C) = (B ∧ C) ∨ A
A ∨ (B ∨ C) = (B ∨ C) ∨ A
They are also distributive, by the following laws:
(A ∧ B) ∨ (A ∧ C)
= A ∧ (B ∨ C)
= (B ∨ C) ∧ A
(A ∨ B) ∧ (A ∨ C)
= A ∨ (B ∧ C)
= (B ∧ C) ∨ A
So many Boolean expressions can be rewritten without parentheses on the right. But, as a counterexample, there is no way to rewrite an expression such as (A ∧ B) ∨ (C ∧ D) if A ≠ C, because of the lack of common factors.
Can't you just do this:
(( A || C ) && D)
and for your second example:
((( A && C ) || B ) && A )
Would that work for you?
Hope that helps...
You'll hit problems if you need to do something like (A || B) && (C || D) unless you can store the intermediate values for later use.
If you're allowed to construct more than one chain and try them all until either one of them passes or they all fail (so each chain is effectively ORed with the next) then I should think you can handle any combination. The example above would become (where each line is a separate query):
(A && C) ||
(A && D) ||
(B && C) ||
(B && D)
However, for a very complex check this could easily get out of hand!