[Math] Find the argument form for the argument and determine whether it is valid

discrete mathematicslogicpredicate-logicpropositional-calculus

I have been trying this question for about 3 hours and I am still stuck please help.

Q: Find the argument form for the argument and determine whether it is valid. Explain your reasoning.

  • If I get a bonus, I will buy the painting.

  • If I sell my books, I will buy the painting.

  • ∴ If I get a bonus or sell my books, I will buy the painting.

We have to build and solve this argument using the rules of inference, however, I have not found a way to do it so far.

What I have done:
List of premises: p –> q, r –> q,
Conclusion: (p V r ) —> q

  1. p –> q (if p then q )
  2. r –> q

    3. 3.

The question should be solved like this example:

Ex: Show that the premises "It is not sunny this afternoon and it is colder than yesterday," "We will go swimming only if it is sunny," "If we do not go swimming, then we will take a canoe trip," and "If we take a canoe trip, then we will be home by sunset" lead to the conclusion "We will be home by sunset."

Solution:

  1. ¬p ^ q (Premise)
  2. ¬p (Simplification using 1)
  3. r –> p (premise)
  4. ¬r (Modus tollens using 2 and 3)
  5. ¬r –> s (Premise)
  6. s (Modus ponens using 4 and 5)
  7. s –> t (Premise)
  8. t (Modus ponens using 6 and 7)

List of available inferences:

modus ponens
modus tollens
hypothetical syllogism
disjunctive syllogism
addition
simplification
conjunction
resolution

We can also use logical equivalences including the ones with conditional statements

Best Answer

$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}$

To use a conditional proof: assume the antecedent aiming to derive the consequent, then (if successful) apply the rule of "Conditional Introduction".

$$\fitch{\text{Premises}}{\fitch{\text{Antecedent}}{\vdots\\\text{Consequent}}\\\text{Antecedent}\to\text{Consequent}}$$

So in this case, you wish to show $p\to q, r\to q \vdash (p\vee r)\to q$.   As the assumption you need to make is a disjunction, you need so rule of inference that will eliminate a disjunction...aka a proof by cases argument.

$$\fitch{1. ~p\to q\\2.~r\to q}{\fitch{3.~p\vee r}{\vdots\\X.~q\qquad\qquad\text{... disjunction elimination}}\\Y.~(p\vee r)\to q\quad\text{3-X,conditional introduction}}$$

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