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I learned that

P, Q1 |- S,  
P, Q2 |- S
--------------------------
P, Q1 | Q2 |- S

Is the converse also true?

P, Q1 | Q2 |- S
--------------------------
P, Q1 |- S

and

P, Q1 | Q2 |- S
--------------------------
P, Q2 |- S

If not, is there a sufficient (introduction) and necessary (elimination) condition for P, Q1 | Q2 |- S?

Thanks.

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  • Yes, if Q1 or Q2 implies S, then Q1 implies S, because Q1 implies Q1 or Q2. Oct 2 at 13:19

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