Question

1. Construct a truth table for: (¬p ∨ (p → ¬q)) → (¬p ∨ ¬q)

2. Give a proof using logical equivalences that
(**p** → **q)** ∨ **(q** →
**r)** and **(p** → **r)**
are ** not** logically equivalent.

3.Show using a truth table that **(p** →
**q)** and **(**¬**q** →
**¬p)** are logically equivalent.

4. Use the rules of inference to prove that the premise
*p* *∧ (p**→*** ¬q)** implies
the conclusion

Answer #1

**Sol.**
**1)**

p | q | ~p | ~q |
p-->~q |
~pV(p-->~q) | ~pV~q | (~pV(p-->~q))-->(~pV~q) |

0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |

0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |

1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |

Table used in this question

Logical Disjunction (OR) (V):

A | B | AVB |

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 1 |

Negation(~)

A | B |

0 | 1 |

1 | 0 |

Logical Implication/if-then (Conditional) (-->)

A | B | A-->B |

0 | 0 | 1 |

0 | 1 | 1 |

1 | 0 | 0 |

1 | 1 | 1 |

Logical conjunction (AND) (/\)

A | B | A /\ B |

0 | 0 | 0 |

0 | 1 | 0 |

1 | 0 | 0 |

1 | 1 | 1 |

**Sol.
2)**

(p-->q) V (q-->r) and (p-->r) are not logically equivalent.

p | q | r | p-->q | q-->r | (p-->q) V (q-->r) | (p-->r) |

0 | 0 | 0 | 1 | 1 | 1 | 1 |

0 | 0 | 1 | 1 | 1 | 1 | 1 |

0 | 1 | 0 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 0 | 1 | 1 | 0 |

1 | 0 | 1 | 0 | 1 | 1 | 1 |

1 | 1 | 0 | 1 | 0 | 1 | 0 |

1 | 1 | 1 | 1 | 1 | 1 | 1 |

From the above clearly appears that (p-->q) V (q-->r) and (p-->r) are not the same. So they are not logically equivalent.

**Sol.**
**3)**

(p-->q) and (~q-->~p) are logically equivalent.

p | q | ~p | ~q | (p-->q) | (~q-->~p) |

0 | 0 | 1 | 1 | 1 | 1 |

0 | 1 | 1 | 0 | 1 | 1 |

1 | 0 | 0 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 1 | 1 |

The above table clearly shows that (p-->q) and (~q-->~p) are logically equivalent.

**Sol.
4)**

p /\ (p-->~q) --> ~q

using rule of modus ponens which states that if p and p-->q is true then we can infer q is true. In tautology it is

rule: p /\ (p-->q) -->q

Therefore

p /\ (p-->~q) --> ~q

We can also prove the above with truth table.

p | q | ~q | p-->~q |

0 | 0 | 1 | 1 |

0 | 1 | 0 | 1 |

1 | 0 | 1 | 1 |

1 | 1 | 0 | 0 |

So you can see when p is true , and (p-->q) is true ~q is also true.

**Note: 0 is also represented as true(T) and 1 is also
represented a false(F).**

are
they logically equivalent (show how) truth table or in word::
a) p —> ( q —> r ) and ( p -> q) —> r
b) p^ (q v r ) and ( p ^ q) v ( p ^ r )

1.Write the negation of the following statement in English. Do
not simply add the words “not” or “it is not the case that” or
similar before the existing statement. “Every dog likes
some flavor of Brand XYZ dog food.”
2. Prove that the compound statements are logically equivalent
by using the basic logical equivalences (13 rules). At each step,
state which basic logical equivalence you are using. (p → q) ∧ (¬r
→ q) and (p ∨ ¬r) → q

Construct a truth table to determine whether the following
expression is a tautology, contradiction, or a contingency.
(r ʌ (p
® q)) ↔ (r ʌ ((r
® p) ®
q))
Use the Laws of Logic to prove the following statement:
r ʌ (p
® q) Û r
ʌ ((r ® p)
® q)
[Hint: Start from the RHS, and use substitution, De Morgan,
distributive & idempotent]
Based on (a) and/or (b), can the following statement be
true?
(p ®
q)...

Use
a truth table to determine whether the two statements are
equivalent.
~p->~q, q->p
Construct a truth table for ~p->~q
Construct a truth table for q->p

1) Show that ¬p → (q → r) and q → (p ∨ r) are logically
equivalent. No truth table and please state what law you're using.
Also, please write neat and clear. Thanks
2) .Show that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology. No
truth table and please state what law you're using. Also, please
write neat and clear.

Using rules of inference prove.
(P -> R) -> ( (Q -> R) -> ((P v Q) -> R) )
Justify each step using rules of inference.

Write a C++ program to construct the truth table of P ||
!(Q && R)

Prove p ∨ (q ∧ r) ⇒ (p ∨ q) ∧ (p ∨ r) by constructing a proof
tree whose premise is p∨(q∧r) and whose conclusion is
(p∨q)∧(p∨r).

(1) Determine whether the propositions p → (q ∨ ¬r) and (p ∧ ¬q)
→ ¬r are logically equivalent using either a truth table or laws of
logic.
(2) Let A, B and C be sets. If a is the proposition “x ∈ A”, b
is the proposition “x ∈ B” and
c is the proposition “x ∈ C”, write down a proposition involving a,
b and c that is logically equivalentto“x∈A∪(B−C)”.
(3) Consider the statement ∀x∃y¬P(x,y). Write down a...

Let P and Q be statements:
(a) Use truth tables to show that ∼ (P or Q) = (∼ P) and (∼
Q).
(b) Show that ∼ (P and Q) is logically equivalent to (∼ P) or (∼
Q).
(c) Summarize (in words) what we have learned from parts a and
b.

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