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.
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