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Originally, when I tried to solve this problem for the first time, my answer was like the following.

1   C Ʌ D       Premiss 
2   C ↔ E       Premiss
3   D ↔ F       Premiss
4   D → F       3 (↔E)
5   C → E       2 (↔E)
6   E Ʌ D       1, 5 
7   E Ʌ F       3, 6

I knew I omitted some steps between some parts of the formulas above, so I tried once again and it became like the following.

1   C Ʌ D       Premiss
2   C ↔ E       Premiss
3   D ↔ F       Premiss
4   D           1 (ɅE)
5   D → F       3 (↔E)
6   F           4, 5 (->E)
7   C           1 (ɅE)
8   C → E       2 (↔E)
9   E           7, 8 (->E)
10  E Ʌ F       6, 9 (ɅI)

As I am not knowledgeable enough to find out every single mistake, I'm wondering if there are any parts to be fixed. Could you let me know any improvement to be made, in case my answer above is not perfect?

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  • 2
    $\begingroup$ Your second attempt looks correct. $\endgroup$ – lemontree Apr 24 at 13:01
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$\begingroup$
 1   C Ʌ D       Premiss
 2   C ↔ E       Premiss
 3   D ↔ F       Premiss
 4   D           1 (ɅE)
 5   D → F       3 (↔E)
 6   F           4, 5 (->E)
 7   C           1 (ɅE)
 8   C → E       2 (↔E)
 9   E           7, 8 (->E)
10   E Ʌ F       6, 9 (ɅI)

Everything is good.

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  • $\begingroup$ Thank you for checking! $\endgroup$ – Shinichi Takagi Apr 30 at 6:34

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