Prove p ∧ (p → q) → q is a tautology using a truth table

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Tek1Atom
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Homework Statement



By providing a truth table, show that the following logic statement is a tautology:

p ∧ (p → q) → qAny help will be much appreciated.

Thank You
 
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p | q | P → q
-------------
T | T | T
T | F | F
F | T | T
F | F | T

Thank you Mark44 but I don't think this is the correct answer/method as the output has to be true in all cases in order for it to be labeled a tautology. Any other suggestions?
 
You are missing at least one column, multiple columns if you want to break things down.
 
D H could you give me an example please as I am new to logic...
 
You have the truth table for implication. p→q obviously is not a tautology. You weren't asked to show that. You were asked to show that p∧(p→q)→q is a tautology. Your truth table needs to end with a p∧(p→q)→q column on the right and with all four values in this column being T.
 
Thank you DH, that was very helpful. My final question is, does it matter which way we write down the truth table?

e.g.

which one is correct?

p | q | P → q | p∧(p→q)→q
---------------------------
T | T | T | T
T | F | F | T
F | T | T | T
F | F | T | T

OR

p | q | P → q | p∧(p→q)→q
---------------------------
F | F | T | T
F | T | T | T
T | F | F | T
T | T | T | T
 
You aren't going to get points for that. At least I hope you aren't. There is a thing called showing your work.
 
Tek1Atom said:
Thank you DH, that was very helpful. My final question is, does it matter which way we write down the truth table?

e.g.

which one is correct?

p | q | P → q | p∧(p→q)→q
---------------------------
T | T | T | T
T | F | F | T
F | T | T | T
F | F | T | T

OR

p | q | P → q | p∧(p→q)→q
---------------------------
F | F | T | T
F | T | T | T
T | F | F | T
T | T | T | T

To answer your question about the order, it doesn't really matter, but these are usually presented with the top row being T T ... and the bottom row being F F ...