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?

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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).

Nice post .Keep sharing

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ReplyDelete{P∨Q,Q→R}⊢¬R→P

{P→Q,Q→R,R→S}⊢P→S