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Showing results for tags 'deduction'.
I don't mean to annoy anyone - i do feel that this is an appropriate (and fun) topic in each of us understanding more what's involved in the conclusion forming process which can either lead to errors in deduction or to progress in our pursuit of accuracy and truth in the solution. one of the real reasons i'm into this thing so much is my passion for 'problem solving,' and i'm sure that's the case for many of ya'll. so these kinds of things are fun, and good for our brains (which we need to solve this thing!) i'm just pasting in this little bit of text and this quick test i found (that I failed) without the answer. if any of you have seen it already, which is very likely, please don't publish the answer, or cheat. so, check it out: If...then... Conditional reasoning is based on an 'if A then B' construct that posits B to be true if A is true. Note that this leaves open the question of what happens when A is false, which means that in this case, B can logically be either true or false. Conditional traps A couple of definitions: In the statement 'If A then B', A is the antecedent and B is the consequent. You can affirm or deny either the antecedent or consequent, which may lead to error. Denying the consequent Denying the consequent means going backwards, saying 'If B is false, then A must also be false.' Thus if you say 'If it is raining, I will get wet', then the trap is to assume that if I am not getting wet then it is not raining. Denying the antecedent Denying the antecedent is making assumptions about what will happen if A is false. Thus if you say 'If it is raining, I will get wet' and is not raining, I might assume that I will not get wet. But then I could fall in the lake. Affirming the consequent This is making assumptions about A if B is shown to be true. Thus if I make the statement 'If it is raining, I will get wet', then if I am getting wet it does not mean that it is raining. The card trap A classic trap was created some years ago; Four cards are laid out as below: The condition is now established (true): 'If a card has a vowel on one side, then it has an even number on the other side.' The problem is to decide which are the minimum cards that need to be turned over to prove that the conditional statement is true. How many and which card(s)? OR (these are the same challenges, just worded a little differently): Which card(s) must you turn over in order to test the truth of the proposition that if a card shows a vowel on one face, then its opposite face is an EVEN number? Discuss it among yourselves...