I have to find out whether this argument is valid or invalid:

p -> q

(q v ([not]r)) -> (p [and] s)s->(r v q)

Hope that makes sense, and also I appreciate any help that can be offered. Thanks.

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- Oct 1st 2009, 07:57 PMGreenDay14Is this argument valid?
I have to find out whether this argument is valid or invalid:

p -> q

__(q v ([not]r)) -> (p [and] s)__s->(r v q)

Hope that makes sense, and also I appreciate any help that can be offered. Thanks. - Oct 4th 2009, 07:41 AMJones
- Oct 4th 2009, 08:22 AMGrandad
Hello GreenDay14I've drawn up a truth table for the proposition

$\displaystyle ((p \rightarrow q)\land((q\lor \neg r)\rightarrow(p\land s)))\rightarrow(s\rightarrow (r\lor q))$

- see attachment.

The columns are evaluated in order (1) through (8) - (8) being the final output. Since this show TRUE in every row, then, yes the argument is valid.

Grandad