I$$ve included line numbers just to make it easier to talk about the different steps.

To give you an idea of what I$$m looking for, here$$s an example of a proof and the kind

of answer I want from you:

Proof.

$1.$ Assume $($P $$ Q$)->$ R and Q $$ P $.$

$2.$ Since $($P $$ Q$)->$ R$,$ we can conclude P $->$ R$.(-$Elim.$)$

$3.$ From Q $$ P $,$ we know P $.(-$Elim.$)$

$4.$ Because we have P $->$ R and P $,$ we know R$.($Appl$.)$

Answer: Line $2$ is wrong because $$ isn$$t the main connective of $($P $$ Q$)->$ R and so

you can$$t use $-$Elim. on it$.$

Finally, note that there are no mistakes in format, phrasing, citations, or other aspects

of presentation, so don$$t worry about that sort of thing. Just pay attention to what

formulas and$/$or subproofs are being used, what rule is being used, what formula is

being concluded, and whether that rule can be used on those formulas$/$subproofs to

deduce that formula.

$($a$)$ Proof.

$1.$ Assume A $->$ C and B$.$

$2.$ Assume A$.$

$3.$ From A and B$,$ we get A $$ B$.(-$Intro.$)$

$4.$ Because A and A $->$ C$,$ we can conclude $$ C $($Appl$.)$

$5.$ Since $$ C$,$ we know C$.($Dbl$.$ Neg.$)$

$6.$ Since we have A $$ B and C$,$ we know $($A $$ B$)->$ C $($Dir$.$ Pf$.)$

$($b$)$ Proof.

$1.$ Assume $($A $$ B$)->($C $$ D$).$

$2.$ Assume B$.$

$3.$ From B$,$ we get A $$ B$.(-$Intro.$)$

$4.($A$$B$)->($C $$D$)$ and A$$B$,$ together imply C $$D$.(->-$Elim.$)$

$5.$ Because C $$ D$,$ C$.(-$Elim.$)$

$6.$ Assuming B$,$ we proved C$,$ and hence B $->$ C$.(->-$Intro.$)$

$($c$)$ Proof.

$1.$ Assume J $$ K and K $->$ L$.$

$2.$ Assume J $->$ L$.$

$3.$ First, consider the case where J is true.

$4.$ From J $->$ L and J$,$ we get L$.($Appl$)$

$5.$ Next, consider the case where K is true.

$6.$ From K $->$ L and K$,$ we get L$.($Appl$)$

$7.$ In either case of J $$ K$,$ we were able to prove L$,$ so

L is true in general.

$($cases$)$

$8.$ Assuming J $->$ L$,$ we proved L$,$ and hence $($J $->$ L$)->$ L$.($dir$.$ pf$)$

$($d$)$ Proof.

$1.$ Assume J $$ K and K $->$ L$.$

$2.$ Case $1$: J$.$

$3.$ Assume J $->$ L$.$

$4.$ From J $->$ L and J$,$ we get L$.($Appl$)$

$5.$ The assumption J $->$L lead to L$,$ and so $($J $->$L$)->$L$.($dir$.$ pf$)$

$6.$ Case $2$: K$.$

$7.$ Assume J $->$ L$.$

$8.$ From K $->$ L and K$,$ we get L$.($Appl$)$

$9.$ The assumption J $->$L lead to L$,$ and so $($J $->$L$)->$L$.($dir$.$ pf$)$

$10.$ In either case of J $$K$,$ we were able to prove $($J $->$L$)->$L$,$

so $($J $->$ L$)->$ L is true in general.

$($cases$)$

$($e$)$ Proof.

$1.$ Assume $($A $$ B$)->$ C and A$.$

$2.$ Case $1$: Assume A$.$

$3.$ From A$,$ we can derive A $$ B$.($Weak$.)$

$4.$ From A $$ B and $($A $$ B$)->$ C$,$ we get C$.($Appl$.)$

$5.$ Case $2$: Assume $$ B$.$

$6.$ From $$ B$,$ we can derive A $$ B$.($Weak$.)$

$7.$ From A $$ B and $($A $$ B$)->$ C$,$ we get C$.($Appl$.)$

$8.$ In either case A or $$ B$,$ we get C$.($Cases$)$

$9.$ From C and A$,$ we can conclude A $$ C$.(-$Intro.$)$

$($f$)$ Proof.

$1.$ Assume $($F $->$ G$)$ G and F $.$

$2.$ Since $($F $->$ G$)$ G is true, so is $$ G$.(-$Elim.$)$

$3.$ From $$ G$,$ we can derive G$.($Dbl$.$ Neg.$)$

$4.$ Because we have F and G$,$ we can conclude F $$ G$.(-$Intro.$)$

$2.$ Some of the following claims are true and some are false. If the claim is true, prove it by

giving a semi$-$formal Natural Deduction proof. If the claim is false, prove this by giving

a truth assignment.

Hint: Start by trying to write a proof. If you get stuck, then switch to trying to find a

proof assignment to disprove the claim.