Another of Tomassi's exercises I can't solve (Logic, page 109, Revision exercise III, 3)
(P v Q) & (R v S) : ((P & R) v (P & S)) v ((Q & R) v (Q & S))
I have to use natural deduction and the only rules I know are:
• assumptions,
• modus ponendo ponens,
• modus tollendo tollens,
• double negation,
• reductio ad absurdum,
• conditional proof,
• v-introduction,
• v-elimination,
• and introduction,
• and elimination.
Tomassi's proof consists of 15 steps.
My answer so far is:
{1} 1. (P v Q) & (R v S) Premise
{1} 2. (P v Q) 1 &E
{1} 3. (R v S) 1 &E
{4} 4. P Assumption for vE
{5} 5. R Assumption for vE
{4,5} 6. P & R 4,5 &I
{4,5} 7. (P & R) v (P & S) 6 vI
{4,5} 8. ((P & R) v (P & S)) v ((Q & R) v (Q & S))
{9} 9. Q Assumption for vE
{10} 10. S Assumption for vE
{9,10} 11. Q & S 9,10 &I
{9,10} 12. ((Q & R) v (Q & S)) 11 vI
{9,10} 13. ((P & R) v (P & S)) v ((Q & R) v (Q & S)) 13 vI
{1,4,9} 14. ((P & R) v (P & S)) v ((Q & R) v (Q & S)) 3,5,8, 10, 13 vE for second conjunct (discharging 5 and 10)
{1,4,5} 15. ((P & R) v (P & S)) v ((Q & R) v (Q & S)) 2,4,8,9,14 vE for first conjunct
What did I get wrong?
Am I allowed to use the conclusion at 14 for both disjuncts of the first disjunction?
Many thanks for your help.
4
,5
,9
or10
?