modus ponens;
$$\begin{array}{|c|c|c|c|c|}\hline\phi &\psi &\phi\Rightarrow\psi&\phi\wedge(\phi\Rightarrow\psi)&[\phi\wedge(\phi\Rightarrow\psi)]\Rightarrow\psi\\hline T&T &T&T&T\ T&F&F&F&T\ F&T&T&F&T\F&F&T&F&T \ \hline \end{array}$$

Question

modus ponens;
$$\begin{array}{|c|c|c|c|c|}\hline\phi &\psi &\phi\Rightarrow\psi&\phi\wedge(\phi\Rightarrow\psi)&[\phi\wedge(\phi\Rightarrow\psi)]\Rightarrow\psi\\hline T&T &T&T&T\ T&F&F&F&T\ F&T&T&F&T\F&F&T&F&T   \ \hline \end{array}$$

unklerhaukus · Answer

@swissgirl

swissgirl · Answer

Its correct  :)

unklerhaukus · Answer

this was the question;

The ancient Greeks formulated a basic rule of reasoning for proving mathematical statements.
Called modus ponens, it says that if you know φ and you know φ ⇒ ψ, then you can conclude ψ. 

(a) Construct a truth table for the logical statement
[φ ∧ (φ ⇒ ψ)] ⇒ ψ 

(b) Explain how the truth table you obtain demonstrates that modus ponens is a valid rule of
inference.

unklerhaukus · Answer

so i have done (a)

im not sure that they want for (b)

unklerhaukus · Answer

do i just have to day that last column is always true? 
 hence [φ ∧ (φ ⇒ ψ)] ⇒ ψ  is always true?

swissgirl · Answer

That is what i was thinking. I am not sure if you have to explain modus ponens
But actually i am remembering from my course that all they wanted was like to say that the truth tables matched blah blah So even though it seems like a stupid answer I have a sneaky feeling that this is what they r looking for

swissgirl · Answer

$  A \to B $        
$  A $      
------
$  ∴B$
Modus ponens is true for all cases as in shown in the truth table, the last column is always true