Chapter 1.6, Problem 12E

Show that the argument form $\left(p\wedge t\right)\to \left(r\vee s\right),q\to \left(u\wedge t\right),u\to p,$and $\neg s$and conclusion $q\to r$is valid by first using Exercise 11 and then using rules of inference from Table 1.

11. Show that the argument form with premisesp₁,p₂, …,p_nand conclusion $q\to r$is valid if the argument form premisesp₁,p₂, …,p_nq,and conclusionris valid.

TABLE 1Rules of Inference.
Rule of Inference	Tautology	Name
$\begin{array}{l}p\\ \therefore \frac{p\to }{q}\end{array}$	$\left(p\wedge \left(p\to q\right)\right)\to q$	Modus ponens

   $\begin{array}{l}\neg q\\ \therefore \frac{p\to q}{\neg p}\end{array}$
   $\left(\neg q\wedge \left(p\to q\right)\right)\to \neg p$ Modus tollens

   $\begin{array}{l}P\to q\\ \therefore \frac{q\to r}{p\to r}\end{array}$
   $\left(\left(p\to q\right)\wedge \left(q\to r\right)\text{}\to \left(p\to r\right)\right)$ Hypothetical syllogism

   $\begin{array}{l}P\vee q\\ \therefore \frac{\neg p}{q}\end{array}$
   $\left(\left(p\vee q\right)\wedge \neg p\right)\to q$ Disjunctive syllogism

   $\therefore \frac{p}{p\vee q}$
   $p\to \left(p\vee q\right)$ Addition

   $\therefore \frac{p\wedge q}{p}$
   $\left(p\wedge q\right)\to p$ Simplification

   $\begin{array}{l}p\\ \therefore \frac{q}{p\wedge q}\end{array}$
   $\left(\left(p\right)\wedge \left(q\right)\right)\to \left(p\wedge q\right)$ Conjunction

   $\begin{array}{l}p\vee q\\ \therefore \frac{\neg p\vee r}{q\vee r}\end{array}$
   $\left(\left(p\vee q\right)\wedge \left(\neg p\vee r\right)\right)\to \left(q\vee r\right)$ Resolution