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i have trouble with understand this type of questions ,

i know how i can determine if the sentences is valid or unsatisfiable in Propositional logic , but in FOL i can't

for example , i have the following sentences in FOL :

1 = 2 

the professor said this is neither valid nor unsatisfiable

but , i think it is always unsatisfiable , because 1 != 2 !!!

can someone explain to me how can i answer this type of question

thank you all

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  • $\begingroup$ It depends on your axioms. Imagine that instead you had the sentence $a = b$ – that would be neither valid nor unsatisfiable. Unless your axioms state the contrary about 1 and 2, then there is no difference between the two statements. $\endgroup$ – Yuval Filmus Jan 13 '17 at 15:50
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A formula is valid if all of the possible truth-assignments to its variables evaluate to true.

A formula is satisfiable if at least one of the truth-assignments to its variables evaluates to true.

The formula you gave is :

$1 = 2$

Since this contains no variables, under any truth-assignment, the expression is $1 = 2$, which is false.

Neither of the above conditions have been met.

Therefore the formula is neither valid nor satisfiable. Since it is not satisfiable, we can say that it is unsatisfiable.

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    $\begingroup$ I'm not sure permutations is the correct word here. $\endgroup$ – Yuval Filmus Jan 13 '17 at 15:49
  • $\begingroup$ Is that better @Yuval Filmus? $\endgroup$ – user3853544 Jan 13 '17 at 15:59
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    $\begingroup$ No, still not the right word. You can say that a sentence is valid if it is satisfied in all interpretations, or for all assignments of its variables. $\endgroup$ – Yuval Filmus Jan 13 '17 at 16:45
  • $\begingroup$ Your statements contradict one another. Permutation also works as it is a super set of the combinations. $\endgroup$ – user3853544 Jan 13 '17 at 17:45
  • $\begingroup$ Combinations implies choosing some unordered subset of a set, and permutations imply choosing an ordered permutation of a set. Truth-assignments is definitely the word you're looking for. $\endgroup$ – jmite Jan 13 '17 at 19:08

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