Translation Guide
If P, then Q (P ⊃ Q)
P, if Q (Q ⊃ P)
What follows “if” is the antecedent of a conditional.
P, provided that Q (Q ⊃ P)
“Provided that” means “if.”
“Assuming that,” “given that,” etc., work the same way. Since they all mean “if,” what follows them is the antecedent of a conditional.
P only if Q (P ⊃ Q)
This says that P is true only if Q is true. So, if P is true, so is Q. Unlike “iff,” it doesn’t tell us that P is false only when Q is false. So,
What follows “only if” is the consequent of a conditional.
P if and only (“iff”) Q (P ≡ Q)
Note that this is equivalent to the conjunction of “P if Q”--i.e., (Q ⊃ P), and “P only if Q”--i.e., (P ⊃ Q).
P unless Q (P v Q) or (~Q ⊃ P)
Think: “P is true, unless Q is,” i.e., “P is true if Q isn’t,” i.e., (~Q ⊃ P). But (~Q ⊃ P) is logically equivalent to (P v Q). While either answer is correct, it is simpler to remember that “unless” just means “or.”
P just if Q (P ≡ Q)
My “ear” tells me “just if” is more like “only if” than “if and only if.” But this is how the author wants it translated, and this is what is expected in Logicola. I won’t put examples of “just if” on the test.
P is sufficient for Q (“enough”) (P ⊃ Q)
P is necessary for Q (Q ⊃ P) or (~P ⊃ ~ Q)
Since P is necessary for Q, Q can’t be true unless P is. So, if Q is true, then it must be that P is also. Hence, “P is necessary for Q” means “If Q, then P.” You might also think: without P, no Q, i.e., (~P ⊃ ~ Q). These are equivalent.
P is necessary and sufficient for Q (P ≡ Q)
P but Q (P . Q)
P or Q, but not both ((P v Q) · ~(P · Q)) or ~(P≡Q)
Not both P and Q ~(P · Q)
Quantified Logic
All As are Bs. (Every A is a B.) (x)(Ax ⊃ Bx)
Everything is A. (x)Ax
Some As are Bs. (At least one A is a B.) (∃x)(Ax · Bx)
Something is A. (∃x)Ax
Nothing is A. (Not even one thing is A.) ~(∃x)Ax log. equiv. to—
--Everything is non-A. (x)~Ax
No As are Bs. (Not even one A is B.) ~(∃x)(Ax · Bx) log. equiv. to— (x)~ (Ax · Bx)
Not all As are Bs. ~(x)(Ax ⊃ Bx) log. equiv. to—
--Some As are not Bs (are non-Bs). (∃x)(Ax · ~Bx) a has property Q. (a is a Q; a Qs) Qa
the so-and-so (this names an object) t (a constant) Something/everything is either A or/and B.
(∃x)(Ax ∨ Bx) (∃x)(Ax • Bx) (x)(Ax ∨ Bx) (x) (Ax • Bx)
--Note that these are quantified statements. But,
Both/Either something/everything is A and/or something/everything is B. ((∃x)Ax v (∃x)Bx) ((∃x)Ax • (∃x)Bx)
((x) Ax v (∃x)Bx) ((x)Ax • (∃x)Bx)
--Note that these are not quantified statements, but disjunctions or conjunctions (the left column are disjunction, the right column conjunctions).
In general:
Quantified statements begin with “(x)” or “(∃x).” (You can apply “drop quantifier rules” only to quantified statements.)
Negated quantified statements begin with “~(x)” or “~(∃x).” (You can apply “reverse squiggle rules” only to negated quantified statements.)
If the statement begins with anything else, you cannot apply “drop universal” or “reverse squiggle” rules to it!
Quantified Translations —With Identity
a is identical to (the same things as) b. a=b or b=a (In general, the order in identity statements doesn’t matter.)
a isn’t identical to b. ~a=b or ~b=a
Bob is a dentist. Db
(“Being a dentist” is a property Bob has.)
Bob is the dentist. b=d
(“The dentist” is a thing that is identical to Bob.)
Someone other than Bob is a dentist. (∃x)(Dx · ~x=b) Someone in addition to Bob is a dentist. (Db · (∃x)(Dx · ~x=b)) Bob alone is a dentist. (No else besides Bob is a dentist.) (Db · ~(∃x)(Dx · ~x=b))
There is at least one dentist. (∃x)Dx
(At least one is … means something is …, but says nothing about whether or not anything else is …)
There is exactly one dentist. (∃x)(Dx · ~(∃y)(~x=y · Dx)) There is something such that it is a dentist and such that there is nothing else (i.e., other than the “first” thing) that is a dentist.
There is something, call it x, such that it is a dentist and such that it is not the case that there is something, call it y, where x isn’t identical to y and y is a dentist.
Or (∃x)(Dx · (y)(Dy ⊃ y=x)) There is an x such x is a dentist and is such that for all y, if y is dentist, then y is identical to x.
There are at least two dentists. (∃x)(∃y)((Dx · Dy) · ~x=y)) (At least two are … means something is … and something not identical to it is …, but says nothing about whether or not anything further is …)
There are exactly two dentists.
(∃x)(∃y)(((Dx · Dy) · ~x=y) · ~(∃z)(Dz · (~z=x · ~z=y)))
There are an x and y such that x and y are dentists, are not identical to one another, and such that there is no z that is a dentist and isn’t identical to either x or y.
Quantified Translations, with Identity and other relations
Romeo loves Juliet. Lrj
Juliet loves Romeo. Ljr (the order matters!)
Romeo gave the flowers to Juliet. Grfj
Someone loves Romeo. (∃x)Lxr
Romeo loves someone. (∃x)Lrx
Everyone loves Romeo. (x)Lxr
Romeo loves everyone. (x)Lrx
No one loves Romeo. ~(∃x)Lxr or (x)~Lxr
Romeo loves no one. ~(∃x)Lrx or (x)~Lrx
All logicians admire Gensler. (x)(Lx ⊃ Axg)
Some logicians admire Gensler. (∃x)(Lx · Axg)
No logician admires Gensler. (Sorry, Harry!) ~(∃x)(Lx · Axg)
Gensler admires all logicians. (x)(Lx ⊃ Agx)
Gensler admires some logicians. (∃x)(Lx · Agx)
Gensler admires no logicians. ~(∃x)(Lx · Lgx)
Some logicians besides Baldner admire Gensler. (∃x)((Lx · ~x=b) · Axg) There are logicians who aren’t Baldner that admire Gensler.
Some logicians in addition to Baldner admire Gensler
(Abg · (∃x)((Lx · ~x=b) · Axg)) No logicians except Baldner admire Gensler.
(Abg · ~(∃x)((Lx · ~x=b) · Axg)) Note on the previous three examples: Do we need to add that Baldner is a
logician (thus adding “ · Lb” somewhere) to the above? This is like the “ … is a boastful druggist” in the previous chapter. It all depends on the context of the argument. Is there anything in the argument as a whole that depends upon the explicit claim that Baldner is a logician? If we wanted to add this clause to the last example, it would come out:
((Lb · Abg) · ~(∃x)((Lx · ~x=b) · Axg))
On the test I will either, like last time, make it clear how I want it translated, or I will accept either (correct!) answer.
— And With Multiple Relations and Mixed Quantifiers!
Someone loves someone. (∃x)(∃y)Lxy
Someone loves everyone. (∃x)(y)Lxy
Everyone loves everyone. (x)(y)Axy
Everyone loves someone.
Note: This might be understood in two distinct ways: “Everyone loves someone or other,”—this is generally what is meant—versus “There is some specific person that everyone loves.” These are not equivalent. Thus:
Everyone loves someone or other. (x)(∃y)Lxy Everyone is such that there is someone they love.
For everyone, there is someone they love.
There is some specific person who everyone loves. (∃x)(y)Lyx There is someone such that everyone loves that person.
There is someone that is loved by everyone.
Pay attention to the difference in the order of “x” and “y” in the above examples! All dentists love one another. (x)(y)((Dx · Dy) ⊃ Lxy) Every banker loves some dentist. (x)(Bx ⊃ (∃y)(Dx • Lxy)) Some banker loves every dentist. (∃x)(Bx • (y)(Dy ⊃ Lxy))
Jones is a lover. (∃x)Ljx
In the text, the author understands “ x is a lover” as “there is someone that x loves.
There is an unloved lover. (∃x)(~(∃y)Lyx • (∃y)Lxy)
There is someone such that no one loves that person, but there is someone that person loves.
For some x, it is false that there is a y such that y loves x and it is true that there is a y such that x loves y.
Everyone loves a lover. (x)((∃y)Lxy ⊃ (y)Lyx)
Everyone loves anybody who loves somebody.
Everyone is such that, if there is someone they love, then they are loved by everyone.
For all x, if there is a y that x loves, then, for all y, y loves x.
Everybody doesn’t love something, but nobody doesn’t love Sara Lee. (an ad slogan from before your time!)
((x)(∃y)~Lxy) . ~(∃x)~Lxs)
It is true both that for all x there is some y such that x doesn’t love y, and there is no x such that x doesn’t love Sara Lee.
