SOLUTION: What is the truth table for this argument? Is this argument valid or invalid? Thank you for helping me!! Rv-R, R&#8594;(F&S), -R&#8594;-(F v S) &#9500; -(F&-S)

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Question 980137: What is the truth table for this argument? Is this argument valid or invalid?
Thank you for helping me!!
Rv-R, R→(F&S), -R→-(F v S) ├ -(F&-S)

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
If F is a truth constant and it represents False, then the conclusion is always true.

F is always false

F & ~S is always false (both pieces need to be true for the conjunction to be true)

~(F & ~S) is always true. It is the complete opposite of the truth value in F & ~S

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Since the conclusion ~(F & ~S) is always true, there is no possible way to have an invalid argument. An invalid argument only occurs if all premises are true with a false conclusion.

This argument is valid. There is no need for a truth table. However, here is a truth table just to be complete.

Premises are in light blue. The conclusion is the last column. The only time you have all true premises is in the last row, but the conclusion isn't false here. This truth table confirms the answer that this argument is valid

SOLUTION: What is the truth table for this argument? Is this argument valid or invalid? Thank you for helping me!! Rv-R, R&#8594;(F&S), -R&#8594;-(F v S) &#9500; -(F&-S)

SOLUTION: What is the truth table for this argument? Is this argument valid or invalid? Thank you for helping me!! Rv-R, R→(F&S), -R→-(F v S) ├ -(F&-S)