Chris J Davis
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Subject says it all, really.
 
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Kārlis Jēriņš
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Sounds like yes, but I could really use an example to be able to give a definitive answer.
 
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Chris J Davis
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I can't remember for sure, but I think the situation was something like this:

If A then C
If C then NOT A
C OR D

Is D true because C causes a contradiction?
 
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Thomas Brendel
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That's not a contradiction. "If A then C" is trivially true when A is false.

A C If A then C
T T T
T F F
F T T
F F T


edit to add: Obviously if the third statement is "A OR D," your question is still relevant. C doesn't cause a contradiction, but A would.
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Jeff Wood
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Remember that IF X THEN Y is equivalent to ~X OR Y.

So your statements become

~A OR C
~C OR ~A
C OR D

Which results in ~A being True, so A is False, all other variables are undefined.
 
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Cameron McKenzie
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On the other hand, if the statement were
A OR (C AND NOT C)
then clearly A would be true, as the part in parenthesis cannot be true.

To prove that a variable is true, you need to be able to demonstrate that assigning it a value of false causes a contradiction regardless of the values of other variables.
In the original variable, you can prove that A is true, because if A were false then you would have a contradiction regardless of the value of C.
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Chris J Davis
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Hmm, okay... Our situation may have been something different then. Unfortunately, I can't think of what it really was now. I just remember that that one side of the OR had been true, it would have caused a paradox, so we played that it meant the other side must be true.

If it comes up again in future games I'll try to make a note of it and post it here.
 
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Cameron McKenzie
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If it gets too confusing, you can always draw up a "truth table"

You basically consider ever possible combination of true and false, and cross out those which are not logically sound based on the premises.

Here this would be my truth table after evaluating "A THEN B"
Basically, I have to mark out every cell where A is true and B is false (because it isn't sound)


A | T T F F
B | T F T F
C D
---
T T X
T F X
F T X
F F X


Maybe next I evaluate "C or A" so I will additionally mark out every cell where C and A are both false:


A | T T F F
B | T F T F
C D
---
T T X
T F X X
F T X
F F X X


Continuing in this way, I get a table that displays all possible combinations of truths.

For complicated statements, I may need to evaluate cells one at a time to determine whether they are sound. That's ok, if it's needed.

If the resulting table is all X's, then the round ends in contradiction.

Otherwise, look at the remaining cells. If A is true in ALL of the remaining cells, then it is proven true. If it is false in all of the remaining cells, then it is proven false. Otherwise, it is indeterminant. Do the same check for B, C, and D.



For example, here is the truth table that results from the following premises:
A THEN NOT B
B AND C
D THEN C

A | T T F F
B | T F T F
C D
---
T T X X X
T F X X X
F T X X X X
F F X X X X


Based on the only two sound conclusion, we know that A is false, B is true, C is true, and D is undetermined.
 
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Chris J Davis
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On a semi-related note, I remember reading somewhere in the rules that you cannot declare Ergo if there is a contradiction in the proof. Does this mean that you cannot declare Ergo if there is a contradiction *anywhere* in the proof, or can you ignore this if the contradiction is proved to be irrelevant? For example, as a result of an IF/THEN clause that doesn't come into effect, or in a statement that doesn't involve your variable? We've been playing the latter way, but wanted to check.
 
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Cameron McKenzie
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bleached_lizard wrote:
On a semi-related note, I remember reading somewhere in the rules that you cannot declare Ergo if there is a contradiction in the proof. Does this mean that you cannot declare Ergo if there is a contradiction *anywhere* in the proof, or can you ignore this if the contradiction is proved to be irrelevant? For example, as a result of an IF/THEN clause that doesn't come into effect, or in a statement that doesn't involve your variable? We've been playing the latter way, but wanted to check.


Fill in the truth table. If you X out everything, then the premises create a contradiction. If there is ANYTHING left in the truth table, then the premises do not form a contradiction.

Note that a statement like "C AND NOT C" will single handedly X out the entire truth table. If any subset of the premises is contradictory, than the entire set of premises is contradictory. There is no scenario in which a contradiction can be resolved by introducing more premises (because the truth table is already X'ed out and further premises cannot remove X's, although changing existing premises can)


Basically, the proof is either contradictory, or it isn't. It can't be partially contradictory. If it seems that the proof is contradictory based on the assumed values of A, B, C, and D, then you have disproven that set of values, but the proof itself may not be contradictory since another set of values may satisfy it.
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