# Math Logic

#### Dumb@Math

##### New member
Can anyone please help me? Im stressing out, ive answered some but I’m still paraniod. An explanation with it will be great too!

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#### Subhotosh Khan

##### Super Moderator
Staff member
Can anyone please help me? Im stressing out, ive answered some but I’m still paranoid. An explanation with it will be great too!
Since you have answered some - please share those with us, so that we know exactly where would we have to start explaining.

#### Dumb@Math

##### New member

I.
1. P
2. X
3. P
4. P
5. P

II.
1. p^¬q
2. ¬p^¬q
3. q⇒(¬r)

but I’m unsure of these answers still

#### Dumb@Math

##### New member
I saw error so this is a better pic of the questions from III to IV

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#### pka

##### Elite Member
I. 1. P 2. X 3. P 4. P 5. P

II.
1. p^¬q 2. ¬p^¬q 3. q⇒(¬r)

but I’m unsure of these answers still
Without knowing the exact way your text defines statement it hard to judge the I above. For example some would say the 4 is an X because it in neither true or false.
In II I agree with 1. & 3. But for 2. $$\neg(p\wedge q)$$.
As you should, this is the implication truth table:
$$\begin{array}{*{20}{c}} p&q&{p \to q} \\ \hline t&t&t \\ t&f&f \\ f&t&t \\ f&f&t \end{array}$$
By looking at that table, consider these two statements.
A true statement is implied by any statement.
A false statement implies any statement.

#### Dumb@Math

##### New member
Without knowing the exact way your text defines statement it hard to judge the I above. For example some would say the 4 is an X because it in neither true or false.
In II I agree with 1. & 3. But for 2. $$\neg(p\wedge q)$$.
As you should, this is the implication truth table:
$$\begin{array}{*{20}{c}} p&q&{p \to q} \\ \hline t&t&t \\ t&f&f \\ f&t&t \\ f&f&t \end{array}$$
By looking at that table, consider these two statements.
A true statement is implied by any statement.
A false statement implies any statement.

This is my answer to some, not yet done tho

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#### pka

##### Elite Member
This is my answer to some, not yet done tho
$$\begin{array}{*{20}{c}} p&q&r&{p \to q}&{\neg p \to r}&{(p \to q) \vee (\neg p \to r)} \\ \hline T&T&T&T&T&T \\ T&T&F&T&T&T \\ T&F&T&F&T&T \\ T&F&F&F&T&T \\ F&T&T&T&T&T \\ F&T&F&T&F&F \\ F&F&T&T&T&T \\ F&F&F&T&F&F \end{array}$$