scala programming language

I have done this solution but required some changes as mentioned below

you have

def isFormula(e: Expr): Boolean =

e match {

case Lit(x) = > x == 0 II x = = 1

case Var(x) => true

case Two("&&", -' e1, e2) = > isFormula(e1) &&


case Three("//", -' e1, e2) = > isFormula(e1) &&



but this is not proper Scala syntax and the case Three is totally wrong as

there are no expressions with 3 operands. The One case is missing too.

it should be something like

def isFormula(e: Expr): Boolean =

e match {

case Lit(x) => x == 0 || x == 1

case Var(x) => true

case Two("&&", op, e1, e2) => isFormula(e1) && isFormula(e2)

case Two("//", op, e1, e2) => isFormula(e1) && isFormula(e2)


So try to add the One case to this and add to the Two cases also the ‘implies' and ‘equiv’.

In the isTautology you have

def isTautology(e: Expr): Option[Boolean] = {

// First of all, check that e represents a formula. If not, then

// no meaningful analysis can be applied, so return None.

if (!isFormula(e))


// At this point e must be a logical expression so the following

// code will determine whether or not it represents a tautology.

e match {

case Lit(x) = > x != 0 II x != 1

case Var(x) => true

case Two("&&", -' e1, e2) = > !isFormula(e1) &&


case Three("//", -' e1, e2) = > !isFormula(e1) &&




but of course this doesn’t make any sense at all as you just pasted

a bit of the isFormula into it and put in a negation in the operands.

So try first to make the isFormula working.

I need this solution within 1 days.

Taidot: Scala, Java

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