AMPS | THARC | KE8QZC | SFW | TSW
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#1:
"If either wages or prices are raised, there will be inflation. If there is inflation, then either Congress must regulate it or the people will suffer. If the people suffer, Congressmen will be unpopular. Congress will not regulate inflation, and Congressmen will not be unpopular. Therefore, wages will not rise."

Solution: Our goal is to conclude $\neg W$:
$\{1\}$(1) $(W \vee P) \rightarrow I$Premise
$\{2\}$(2) $I \rightarrow C \vee S$Premise
$\{3\}$(3) $S \rightarrow U$Premise
$\{4\}$(4) $\neg C \wedge \neg U$Premise
$\{4\}$(5) $\neg U \wedge \neg C$$\wedge$-commutativity on 4
$\{4\}$(6) $\neg C$$\wedge$-elimination on 4
$\{4\}$(7) $\neg U$$\wedge$-elimination on 5
$\{3\}$(8) $\neg U \rightarrow \neg S$contrapositive of 3
$\{3,4\}$(9) $\neg S$$\rightarrow$-elimination on 7 and 8
$\{2\}$(10) $\neg(C \vee S) \rightarrow \neg I$contrapositive of 2
$\{1\}$(11) $\neg I \rightarrow \neg (W \vee P)$contrapositive of 1
$\{1,2\}$(12) $\neg (C \vee S) \rightarrow \neg (W \vee P)$syllogism on 10 and 11
$\{3,4\}$(13) $\neg C \wedge \neg S$$\wedge$-introduction on 6 and 9
$\{3,4\}$(14) $\neg (C \vee S)$DeMorgan's law on 13
$\{1,2,3,4\}$(15) $\neg(W \vee P)$$\rightarrow$-elimination on 12 and 14
$\{1,2,3,4\}$(16) $\neg W \wedge \neg P$DeMorgan's law on 15
$\{1,2,3,4\}$(17) $\neg W$$\wedge$-elimination on 16


#2:
"Either logic is difficult, or not many students like it. If mathematics is easy, then logic is not difficult. Therefore, if many students like logic, mathematics is not easy."

Solution: Our goal is to conclude $L \rightarrow \neg M$:
$\{1\}$(1) $D \vee \neg L$Premise
$\{2\}$(2) $M \rightarrow \neg D$Premise
$\{1\}$(3) $\neg L \vee D$$\vee$-commutativity on 1
$\{1\}$(4) $L \rightarrow D$equivalence of implication and disjunction on 3
$\{2\}$(5) $\neg (\neg D) \rightarrow \neg M$contrapositive of 2
$\{6\}$(6) $D$Premise
$\{6\}$(7) $\neg(\neg D)$double negation of 6
$\{2,6\}$(8) $\neg M$$\rightarrow$-elmination on 5 and 7
$\{2\}$(9) $D \rightarrow \neg M$rule of conditional proof on 6 and 8
$\{1,2\}$(10) $L \rightarrow \neg M$syllogism on 4 and 9


#3:
"If Algernon is in jail, then he is not a nuisance to his family. If he is not in jail, then he is not a disgrace. If he is not a disgrace, then he is in the army. If he is drunk, he is a nuisance to his family. Therefore, he is either not drunk or in the army."

Solution: Our goal is to conclude $\neg R \vee A$. To do it, we will first conclude $R \rightarrow A$ and then apply the equivalence of implication and disjunction:
$\{1\}$(1) $J \rightarrow \neg N$Premise
$\{2\}$(2) $\neg J \rightarrow \neg D$Premise
$\{3\}$(3) $\neg D \rightarrow A$Premise
$\{4\}$(4) $R \rightarrow N$Premise
$\{1\}$(5) $\neg (\neg N) \rightarrow \neg J$contrapositive of 1
$\{6\}$(6) $N$Premise
$\{6\}$(7) $\neg (\neg N)$double negative of 6
$\{1,6\}$(8) $\neg J$$\rightarrow$-elimination on 5 and 7
$\{1\}$(9) $N \rightarrow \neg J$rule of conditional proof on 6 and 8
$\{1,4\}$(10) $R \rightarrow \neg J$syllogism on 4 and 9
$\{1,2,4\}$(11) $R \rightarrow \neg D$syllogism on 2 and 10
$\{1,2,3,4\}$(12) $R \rightarrow A$syllogism on 3 and 11
$\{1,2,3,4\}$(13) $\neg R \vee A$law of equivalence of implication and disjunction on 12


#4:
"Either John and Henry are the same age, or John is older than Henry. If John and Henry are the same age, then Elizabeth and John are not the same age. If John is older than Henry, then John is older than Mary. Therefore, either Elizabeth and John are not the same age or John is older than Mary."
Solution: The goal is to derive $\neg E \vee M$. We will do so by deriving $E \rightarrow M$ and applying the Law of Equivalence of Implication and Disjunction:
$\{1\}$(1) $S \vee O$Premise
$\{2\}$(2) $S \rightarrow \neg E$Premise
$\{3\}$(3) $O \rightarrow M$Premise
$\{2\}$(4) $\neg(\neg E) \rightarrow \neg S$contrapositive of 2
$\{5\}$(5) $E$Premise
$\{5\}$(6) $\neg (\neg E)$double negative of 5
$\{2,5\}$(7) $\neg S$$\rightarrow$-elimination of 4 and 6
$\{2\}$(8) $E \rightarrow \neg S$conditional proof on 5 and 7
$\{9\}$(9) $\neg S$Premise
$\{1,9\}$(10) $O$modus tollendo tollens on 1 and 9
$\{1\}$(11) $\neg S \rightarrow O$conditional proof on 9 and 10
$\{1,2\}$(12) $E \rightarrow O$syllogism on 8 and 11
$\{1,2,3\}$(13) $E \rightarrow M$syllogism on 3 and 12
$\{1,2,3\}$(14) $\neg E \vee M$equivalence of implication and disjunction on 13