For example, given the valid formula $\forall x(Rxx \rightarrow \exists y Rxy)$, it gives the following tableau proof:
1. $\neg \forall x(Rxx \rightarrow \exists y Rxy)$
2. $\neg (Raa \rightarrow \exists yRay)$
3. $Raa$
4. $\neg \exists yRay$
5. $\neg Raa$
Shouldn't the last line (5) read "$\neg Raa$"?
ReplyDeleteOops - yeah!
ReplyDeleteI'm trying to implement an automated tree proof generator for formulae of propositional logic. Do you know any sources where one could find tips on how to implement a proof generator like the one you've linked to?
ReplyDeleteNERD!!
ReplyDeleteCan you help me figure this problem out?
ReplyDelete:((P --> Q) V (Q --> R))
1. $\neg (P \rightarrow Q) \vee (Q \rightarrow R))$.
ReplyDelete2. $\neg(P \rightarrow Q)$. (from (1), by $\neg \vee$-rule)
3. $\neg(Q \rightarrow R)$. (from (1), by $\neg \vee$-rule)
4. $P$. (from (2), by $\neg \rightarrow$-rule)
5. $\neg Q$. (from (2), by $\neg \rightarrow$-rule)
6. $Q$. (from (3), by $\neg \rightarrow$-rule)
7. $\neg R$. (from (3), by $\neg \rightarrow$-rule)
X
The tableau is closed, as (5) contradicts (6).
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ReplyDelete