Question #238877

Show that p ⋁ (q ⋀ r) and (p ⋁ r) ∧ (p ⋁ r) are logically equivalent. This

is the distributive law of disjunction over conjunction.


Expert's answer

Distributive law of disjunction over conjunction has the form

p∨(q∧r)=(p∨q)∧(p∨r)p\lor(q\land r)=(p\lor q)\land(p\lor r)

Let we prove this identity

We must conider all cases for p,q,r∈⦃0,1⦄\in \lBrace 0,1\rBrace

1) Let p=1

Left part of the equation has the form

1∨(p∧r)=11\lor(p\land r)=1 because 1∨X=11\lor X=1 for any X. be the properties of 1.

Right part equals to(1∨q)∧(1∨r)=1∧1=1(1∨ q)\land (1\lor r)=1\land 1=1

Thus if p=1 both parts of equation eqaul to 1 therefore the identity is true

for all possible values q,r∈⦃0,1⦄\in \lBrace 0,1 \rBrace .

Now let be p=0.

In this case left part of identity equals to 0∨(q∧r)=q∧r0\lor(q\land r)=q\land r

because 0∨X=X0\lor X=X by the properties of 0.

Right part equals to (0∨q)∧(0∨r)(0\lor q)\land(0\lor r) =q∧rq\land r

and q∧r≡q∧rq\land r \equiv q\land r

Thus in all possible cases p=0 and p=1 left and right parts are equal identically, so identity is proved.

Given equality p ⋁ (q ⋀ r) and (p ⋁ r) ∧ (p ⋁ r) contans mistake because if p=0,r=1 right part of it equals (0∨1)∧(0∨1)=1∧1=1(0\lor 1)\land (0\lor 1)=1\land 1=1

(0∨1)∧(0∨1)=1∧1=1(0\lor 1)\land (0\lor 1)=1\land 1=1 but left part eqals to

0∨(q∧1)=0∨q=q0\lor(q\land 1)=0\lor q=q

and we must have q=1 but this is not necessary and may be q=0 and so given identity is erroneous.


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