By Alain Prouté

Un polynôme doit-il se noter P ou P(X) ? Réponse dans ce texte.

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So we added and ~p. Then, we process the conjunction: Conjunctions are straightforward: you ensure that both conjuncts are true. So, we add and q to our tree. Finally, we process the conditional and for this we ensure either that ~p is true (the left branch) or that q is true (the right branch). Both branches of the tree stay open, and we indicate this with the vertical arrow at the bottom of the branches. Both branches are said to be completed, and both are open. Therefore, both represent ways to satisfy the formula Select a branch (I have chosen the left one), and read up that branch to find each atomic formula occurring by itself in the branch.

Therefore, there is no evaluation satisfying all of X and also ~A, so X, ~A╞. Conversely, if X, ~A╞ then there is no evaluation making the premises X true and the conclusion A false, and so the argument is valid. We have X╞A. The idea behind trees The method for trees goes like this: to test an argument, put the premises of the argument and the negation of the conclusion in a list. We want to see if this list can be satisfied—if the propositions cannot be true together, the argument is valid; if they can be true together, the argument is not valid.

These are each satisfied by I, since that is what we assumed. Now, for each rule that is applied to extend the tree, at least one of the branches generated will give us formulas that are satisfied by I, provided that the formula resolved is also satisfied by I. We take the rules one at a time. If we resolve a double negation ~~A then we know that I(~~A)= 1. The rule adds A to the branch, but we can see that I(A)=1 too. If we resolve a conjunction A&B then we know that I(A&B)= 1. We add A and B to the branch, but clearly I(A)=1 and I(B)=1, so these formulas are satisfied too.

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