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Translate this sentence in predicate logic : "Everyone loves her or his parents, and everyone has two parents. Therefore, everyone loves at least two people." Domain = the set of all people ; Fx = x is a female ; Mx = x is a male ; Lxy = x loves y ; Pxy = x is a parent of y. Is the argument valid (i.e. on every interpretation, if the first sentence is true, the second is true)?


My guess is : ((∀x)(∀y)(((Fx ∨ Mx) ∧ Pyx) ⊃ Lxy) ∧ (∀x)(∃y)(∃z)((Pyx ∧ Pzx) ∧ ¬ y=z)) ⊃ (∀x)(∃y)(∃z)(¬ y=z ∧ (Lxy ∧ Lxz)). I think it is invalid, but this is only if we specify, like I did, (Fx ∨ Mx).

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  • $\begingroup$ This is the interpretation under which the sentence is invalid : Domain = the set of all integers ; Fx = x is negative ; Mx = x is inferior to 1 ; Pxy = x is greater that x ; Lxy = x is the square root of y. $\endgroup$ – Thrasymaque May 11 '17 at 1:50
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    $\begingroup$ Unless you cannot use equality, $x = y$, you do not need to mention anything about gender. It is not used in the statements which you are translating. (Other than the "his or her" pronoun, which is a fish of the wrong colour) $\endgroup$ – Graham Kemp May 11 '17 at 1:59
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    $\begingroup$ It's not a sentence, but an argument involving multiple sentences. $\endgroup$ – Bram28 May 11 '17 at 2:03
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    $\begingroup$ Just take out the $Fx \lor Mx$ part, split it into three separate sentences, and you're all good! $\endgroup$ – Bram28 May 11 '17 at 2:05
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This is not a sentence, but an argument (as the later question 'Is the argument valid?' also indicates) involving multiple sentences. So you really shouldn;t symbolize this with a conditional, but instead have three separate sentence, 2 premises and 1 conclusion.

Also, you can take out the $Fx \lor Mx$ part: it says 'everyone', so it should be everyone, and if you are concerned about gender-neutral persons, this sentence clearly is not, since it continues with 'his or her'. So yes, you would be right that this would be invalid if only "he"'s and "she"'s have parents, and that there might be people who are not either "he" or "she', but the sentences clearly assumes everyone to be a "he" or "she". So: you can either add the $Fx \lor Mx$ part to every sentence, or (which would be my preference, since with the assumption that everyone is a "he" or "she", the $Fx \lor Mx$ part ends up saying nothing of interest) leave it out completely.

So, this leaves you with:

$\forall x \forall y (Pyx \rightarrow Lxy)$

$\forall x \exists y \exists z (Pyx \land Pzx \land y \not = z)$

$\therefore \forall x \exists y \exists z (y \not = z \land Lxy \land Lxz)$

... which is easily shown to be valid.

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