Question

In: Advanced Math

FOR EAICH PAIR OF PROPOSITIONS P AND Q STATE WHETHER ON NOT p=q p=(s→(p ∧¬r)) ∧...

FOR EAICH PAIR OF PROPOSITIONS P AND Q STATE WHETHER ON NOT p=q

p=(s→(p ∧¬r)) ∧ ((p→(r ∨ q)) ∧ s), Q=p ∨ t

Solutions

Expert Solution

Truth tabl

                           P       Q
   p     q    r     s    t    ~r p^~r s->(p^~r)     rVq p->(rVq) (p->(rV q))^s      (s->(p^~r))^((p->(rVq))^s)    pVt
T    T    T    T    T    F T T T T T T T
T    T    T    T    F    F F F T T T F T
T    T    T    F    T    F T T T T F F T
T    T    T    F     F    F F T T T F F T
T    T    F    T    T    T T T T T T T T
T    T    F    T    F    T F F T T T F T
T    T    F     F    T    T T T T T F F T
T    T    F     F    F    T   F T T T F F T
T    F T     T    T    F T T T T T T T
T    F T     T F    F F F T T T F T
T    F T     F    T     F T T T T F F T
T    F T     F    F     F F T T T F F T
T    F    F    T    T    T T T F F F F T
T    F    F    T    F     T F F F F F F T
T    F    F     F     T    T T T F F F F T
T    F    F    F    F     T F T F F F F T
F    T    T    T    T     F F F T T T F T
F    T    T    T    F     F F F T T T F F
F    T    T    F    T    F F T T T F F T
F    T    T    F F     F F T T T F F F
F    T    F    T     T     T F F T T T F T
F    T    F    T    F     T F F T T T F F
F    T    F    F    T     T F T T T F F T
F    T    F    F     F     T F T T T F F F
F    F    T    T    T     F F F T T T F T
F    F    T    T    F     F F F T T T F F
F    F    T     F    T     F F T T T F F T
F    F    T    F    F     F F T T T F F F
F    F    F    T    T    T F F F T T F T
F    F    F    T    F    T F F F T T F F
F    F    F    F    T    T F T F T F F T
F    F    F    F    F    T F T F T F F F

fBy analyzing truth table we can conclude that P != Q.


Related Solutions

Prove p → (q ∨ r), q → s, r → s ⊢ p → s
Prove p → (q ∨ r), q → s, r → s ⊢ p → s
The following defines what logical property: A pair of propositions (wffs) P,Q that have the same...
The following defines what logical property: A pair of propositions (wffs) P,Q that have the same truth value under every interpretation. consistency? proposition? argument? equivalence ?
Suppose S = {p, q, r, s, t, u} and A = {p, q, s, t}...
Suppose S = {p, q, r, s, t, u} and A = {p, q, s, t} and B = {r, s, t, u} are events. x p q r s t u p(x) 0.15 0.25 0.2 0.15 0.1 (a) Determine what must be p(s). (b) Find p(A), p(B) and p(A∩B). (c) Determine whether A and B are independent. Explain. (d) Arer A and B mutually exclusive? Explain. (e) Does this table represent a probability istribution of any random variable? Explain.
Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.)...
Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.) Why does this equivalency allow us to use the proof by contradiction technique?
Show that if P;Q are projections such that R(P) = R(Q) and N(P) = N(Q), then...
Show that if P;Q are projections such that R(P) = R(Q) and N(P) = N(Q), then P = Q.
A list of six positive integers, p, q, r, s, t, u satisfies p < q...
A list of six positive integers, p, q, r, s, t, u satisfies p < q < r < s < t < u. There are exactly 15 pairs of numbers that can be formed by choosing two different numbers from this list. The sums of these 15 pairs of numbers are: 25, 30, 38, 41, 49, 52, 54, 63, 68, 76, 79, 90, 95, 103, 117. Which sum equals r + s?
Discrete math question Prove that ¬(q→p)∧(p∧q∧s→r)∧p is a contradiction without using truth table
Discrete math question Prove that ¬(q→p)∧(p∧q∧s→r)∧p is a contradiction without using truth table
Two compound propositions p and q in propositional logic are logically equivalent if . . ..
Complete the following statements.Two compound propositions p and q in propositional logic are logically equivalent if . . ..An argument form in propositional logic is valid if . . ..A theorem is a statement that . . ..A statement that is assumed to be true is called a(n) . . ..A proof is a valid argument that . . ..
Q1) Determine whether each of the compound proposition is satisfiable. (p ∨ q ∨ r) ∧...
Q1) Determine whether each of the compound proposition is satisfiable. (p ∨ q ∨ r) ∧ (p ∨ ¬q ∨ ¬s) ∧ (q ∨ ¬r ∨ s) ∧ (¬p ∨ r ∨ s) ∧ (¬p ∨ q ∨ ¬s) ∧ (p ∨ ¬q ∨ ¬r) ∧ (¬p ∨ ¬q ∨ s) ∧ (¬p ∨ ¬r ∨ ¬s) Q2) Negate the following statements, no negation symbol before quantifiers. 1) ∃x∃y(Q(x, y) ↔ Q(y, x)) 2) ∀y∃x∃z(T (x, y, z) ∨ Q(x,...
Let p and q be the propositions p:You drive over 65 miles per hour. q: You...
Let p and q be the propositions p:You drive over 65 miles per hour. q: You get a speeding ticket. Write these propositions using p and q and logical connectives. a) You do not drive over 65 miles per hour. b) You drive over 65 miles per hour, but you do not geta 1 speeding ticket. c) You will get a speeding ticket if you drive over 65 miles per hour. d) If you do not drive over 65 miles...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT