Thread: help me

please help me to slove that.

Question 1 is wrong, unless :r means not r.

If r is true (the way I see it) then p, q being false still keep the statement true.

If r is false (which I am assuming) then both p,q cannot be false because that would imply p v (q v r)=FALSE
Thus, one of p,q is true
Thus, p v q=TRUE

Hello, m777!

I assume you are familiar with these definitions, properties and axioms.

p → q . ↔ . ~p V q . . . . . . . .Definition of Implication

p V q . ↔ . q V p . . . . . . . . . Commutative property of Disjunction

p V (q V r) . ↔ . (p V q) V r . . Associative property of Disjunction

p V ~p . → . t . . . . . . . . . . . Axiom #1

t V p . → . t . . . . . . . . . . . . .Axiom #2

3(b) Deterine whether the statement is a tautology, a contingency, or absurdity.

. . p → (q → p)

We have: .p → (q → p)

. . . . . . . .p → (~q V p) . . . .Def. of Implication

. . . . . . . ~p V (~q V p) . . . Def. of Implication

. . . . . . . (~p V p) V ~q . . . Comm/Assoc properties

. . . . . . . . . . t V ~q . . . . . .Axiom #1

. . . . . . . . . . . .t . . . . . . . . Axiom #2


The statement is always true . . . It is a tautology.


(A truth table would have been faster . . . )