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# Chapter+8+Review+Sheet - Review Sheet for Exam 4...

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Unformatted text preview: Review Sheet for Exam 4: Derivations in Predicate Logic: All the rules and derivation techniques that we used in sentential logic (exam 2) can also be used for predicate logic derivations (exam 4). So refresh your memory on those. There are five new rules and one new derivation technique. New rules: In the following rules we use the following symbolizations: v is any variable (x,y,z); n is a name (a,b,c etc.). F[v] is a formula. F[n] is the formula that results when n replaces v at all free occurrences of v in F[v]. ∀O (universal-out): ∀vF[v] F[n] Here, n can be any name. ƎO (existential-out): ƎvF[v] F[n] Here, n must be a NEW name (that is, n cannot have appeared on any previous line of the proof- including in the formula upon which you are using the rule). NOTICE: Universal-out and existential-out are very similar: in fact the only difference is that in existential-out one has to use a new name, but in universal-out one can use any name. ƎI (existential-in) F[n] ƎvF[v] Here, n can be any name. ~∀O (tilde-universal-out): ~∀vF[v] Ǝv~F[v] ~ ƎO (tilde-existential-out): ~ ƎvF[v] ∀v~F[v] Notes: Free and bound variables: A variable is free if and only if it is not bound by a quantifier. A quantifier binds a variable when both (a) the variable is governed by that quantifier, and also (b) the variable is within the scope of the quantifier. A quantifier governs the variable to which it is attached. So, in ∀x the universal governs x, in ∀y the universal governs y, and in ∀z the universal governs z (similar comments apply when the connective is the existential). The scope of a quantifier depends on what part of the formula the quantifier is applying to. You can tell this largely by looking at the position of parentheses. Its use is a little like in math. Consider the following equations from math: (1) 3(x+y) – z (2) 3x + (y-z) (3) 3[(x+y)-z] (4) (x+y) – 3z In (1), 3 applies only to x and y, but not z. One could say that x and y are in the scope of 3, but z isn’t. In (2), only x is in the scope of 3, but y and z aren’t. In (3), x, y, and z are all in the scope of 3. In (4), only x is in the scope of 3. Similarly in logic (in the case here y is attached to the quantifier so it governs ys only): (1) ∀y(Fy & Gy) Hy (2) ∀yFy & (Gy Hy) (3) ∀y[(Fy & Gy) Hy] (4) (Fy & Gy) ∀yHy In (1), ∀y applies only to the ys in Fy and Gy, but not that in Hy (Fy and Gy are in the scope of ∀y but Hy is not). In (2), only Fy is within the scope of ∀y. In (3), Fy, Gy and Hy are all within the scope of ∀y. In (4), only Hy is within the scope of ∀y So, from this we know that in (1) the ys in Fy and Gy are bound by ∀y, while the y in Hy is not bound- it is free. In (2), the y in Fy is bound by ∀y, but the occurrences of y in Gy and Hy are free. And in (3), all occurrences of y are bound (all of them are within the scope of and governed by ∀y). In (4), the y in Hy is bound, but those in Fy and Gy are free. Let’s see how this information matters when it comes to the rules. This discussion is most relevant to the use of universal-out. Recall that universal-out allows one only to replace free occurrences of v in F[v]. Make sure you are clear what F[v] actually is- it is the formula that results when one removes the universal quantifier from the front. Examples: (1) Suppose I have available the formula ∀yFy. I know I can use ∀O because the main connective is the universal. However, I need to discover whether I can replace the y in Fy with a name (any name for this rule). The rule is written such that we start with ∀vF[v], and we substitute free occurrences of v in F[v] with a name. Here ∀vF[v] is ∀yFy, so F[v] is Fy. So, I need to check that the occurrence of y in Fy is free. Well it clearly is (since there is no quantifier here at all it is easy to figure this out). So we can replace the y in Fy with any name we want. E.g. we could get Fa. Or we could get Fb etc. (2) Suppose I have available the formula ∀y(Fy Gy). Here my ∀vF[v] is ∀y(Fy Gy), so my F[v] is FyGy. I check to see whether the ys in FyGy are free. Both of them are free (neither are bound by a quantifier) so I can replace both. Hence I could get FaGa. Or I could get FbGb etc. (3) Suppose I have available the formula ∀x(Fx ∀x Gx). Here my ∀vF[v] is ∀x(Fx ∀x Gx). So my F[v] is Fx∀x Gx. I check to see if the x in Fx and that in Gx are free. Well the x in Fx is free, so I can replace that with any name. However, the x in Gx is not free since the Gx is governed by ∀x (the one in F[v]). So I can only replace the x in Fx with a name; I can’t alter the one in Gx. Hence, I can get Fa∀x Gx. Or I can get Fb∀x Gx etc. (4) Suppose I have available the formula ∀xƎy[Fx ∀y(Gy & Hx)]. Here my F[v] is going to be Ǝy[Fx ∀y(Gy & Hx)]. I check to see if the x in Fx and that in Hx are free. Notice that the y in Gy is irrelevant here, since the universal out deals with ∀x, which governs x (and only x). While Fx is within the scope of a quantifier, that quantifier does not govern x (since Ǝy governs y only). Hence, the x in Fx is free. While Hx is within the scope of two quantifiers (Ǝy and ∀y), both these quantifiers govern y, and so do not bind x. Hence, the x in Hx is free. Hence, using the rule we can get Ǝy[Fa ∀y(Gy & Ha)]. Or we can get Ǝy[Fb ∀y(Gy & Hb)] etc. New Derivation Technique: Universal-derivation. This technique can only be used when one needs to show a universal formula (a formula whose main connective is a universal quantifier). In the following v is any variable (x,y,z); n is any name (a,b,c etc.). F[v] is a formula. F[n] is the formula that results when n replaces v at all free occurrences of v in F[v]. SHOW: ∀vF[v] | SHOW: F[n] || || Here, n must be a new name. Notice that the formula that we need to show here is one that could be obtained from ∀vF[v] using universal-out- although in addition (and unlike in universal-out) we need to check to make sure that the name we use is new. However, don’t confuse this derivation technique with a rule. These lines are not available for use (at least until one has been able to put a line through the show). Notice also that this derivation technique is different from the others we have seen (DD, ID and CD) in that it requires that there is a second show line immediately after the first one. As said, one has to introduce a new name. Notice that one cannot use any name that has appeared anywhere in the proof so far, including on any of the showlines. So suppose that I have to SHOW: ∀x(FxGa) My next line (the showline) can therefore NOT use an ‘a’ since this letter appears in the formula ∀x(FxGa). So my showline could be: SHOW: FbGa, or SHOW: FcGa E.g. SHOW: ∀x(FxGa) | SHOW: FbGa Tips on how to do derivations: The type of derivation you use will depend on the main connective of the formula you need to show. The same tips apply as were given for sentential logic where the main connective is one of our original set e.g. where the main connective is an arrow, we would use conditional derivation. Here are tips concerning the new cases: (1) Where you have to show an existential formula, ƎvF[v]: (i) indirect derivation will always work; (ii) direct derivation will sometimes work. e.g. using ID: assume the negation of the formula, and then show an X. In addition here, immediately use the ~ ƎO rule on the assumption line to give yourself a useful formula to work with. SHOW: ƎvF[v] ID | ~ ƎvF[v] | SHOW: X | | ∀v~F[v] As DD ~ ƎO (on assumption line). e.g. using DD. To do a direct derivation you will probably have to find a formula upon which you can use existential-in to get to your conclusion. This will not always be possible so you are taking a bit of a risk (a risk of wasting time that is) if you do this technique. (2) Where you have to show a universal formula, ∀vF[v]: (i) universal derivation will always work and will be the easiest method to use; (ii) indirect derivation will always work but will probably be a lot harder. (3) Where you have to show a formula of type ~ ƎvF[v] or ~∀vF[v] then you should ALWAYS use indirect derivation. You will assume the un-negated version of the formula and then find an X. E.g. 1: SHOW: ~ ƎvF[v] | ƎvF[v] | SHOW: X || ID As DD E.g. 2: SHOW: ~∀vF[v] | ∀vF[v] | | SHOW: X || ID As DD Tips about the use of rules. (1) If you have available a formula of type ~ ƎvF[v], then immediately use ~ ƎO. This will produce a universal statement that will be useful at some point. (2) If you have available a formula of type ~∀vF[v], then immediately use ~∀O. This will produce an existential statement that will be useful at some point. (3) [Rule of thumb only] If you have a choice between using existential-out or universal-out, use existential out first. (4) [Rule of thumb only] If no names have yet appeared in the proof (either on available lines or showlines) then do not yet use universal-out (if you could anyway of course!) A hardish example: this is 68 in the book (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝ y(Fy & ~ Ryx)] Pr I have to show a universal formula so I use universal derivation. My next line will be a showline. I can figure out what this line will be by pretending that I am using universalout on ∀x[Gx Ǝ y(Fy & ~Ryx)], and then checking that the name I introduce is a new name. Using universal-out on ∀x[Gx Ǝ y(Fy & ~Ryx)] would give me Ga Ǝ y(Fy & ~Rya). I check and see that the letter ‘a’ has not been used yet so I can use it here. So I now know my next showline. (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝ y(Fy & ~Ryx)] (3) | SHOW: Ga Ǝ y(Fy & ~ Rya) Pr UD I see that I have to show a conditional statement so now I use conditional derivation: (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝ y(Fy & ~Ryx)] (3) | SHOW: Ga Ǝ y(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝ y(Fy & ~ Rya) Pr UD CD As I now need to show an existential statement. My tips said that ID would always work, while DD may or may not work. To play it safe then I will use ID here. Notice that DD does in fact work here. The book uses DD here and so you can check that way of doing it in the back of the book (it’s question 68 in the book). (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga Pr UD CD As (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~ Rya) (7) | | | SHOW: X ID As DD My tip suggested to use ~ ƎO immediately when one sees a formula of type ~ƎvF[v]. We have a formula like this on line 6, so I will use ~ ƎO on line 6. (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) Pr UD CD As ID As DD 6,~ ƎO Now, lines 1, 4, 6 and 8 are available. I’ve just used line 6 so I won’t need that again. I have an existential statement and a universal one (lines 1 and 8 respectively). My rule of thumb says to do existential-out before universal out. So I will use existential-out on line 1 next. Notice that in doing existential-out I have to introduce a new name. Hence, I cannot use ‘a’ since this has appeared in the proof already. Hence I will use ‘b’, which is new. Notice also that I will be replacing both occurrences of x in Fx & ~ Ǝ y(Gy & Rxy) with the name. This is because both xs in this formula are free. (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) Pr UD CD As ID As DD 6,~ ƎO 1, ƎO To make it clearly where to go now it’s worth breaking up line 9 using ampersand-out (twice). (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) Pr UD CD (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) (10)| | | | Fb (11)| | | | ~ Ǝ y(Gy & Rby) As ID As DD 6,~ ƎO 1, ƎO 9,&O 9,&O I see now that I have a formula of type ~ƎvF[v] so I know to use ~ƎO immediately. (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) (10)| | | | Fb (11)| | | | ~ Ǝ y(Gy & Rby) (12)| | | | ∀y~(Gy & Rby) Pr UD CD As ID As DD 6,~ ƎO 1, ƎO 9,&O 9,&O 11,~ ƎO Now, notice that line 4 is Ga and Gy appears in line 12. I know then that when I use universal-out on line 12, I should use an ‘a’. Notice also that line 10 is Fb and that Fy appears in line 8. I know then that when I use universal-out on line 8 I should use ‘b.’ I do both of these things now: (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) (10)| | | | Fb (11)| | | | ~ Ǝ y(Gy & Rby) Pr UD CD As ID As DD 6,~ ƎO 1, ƎO 9,&O 9,&O (12)| | | | ∀y~(Gy & Rby) 11,~ ƎO (13)| | | | ~(Ga & Rba) 12, ∀O (14)| | | | ~(Fb & ~Rba) 8, ∀O I am now in a position to use the rule ~&O. In fact I can use this on both lines 13 and 14 (separately of course). I do these two steps now: (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) (10)| | | | Fb (11)| | | | ~ Ǝ y(Gy & Rby) (12)| | | | ∀y~(Gy & Rby) (13)| | | | ~(Ga & Rba) (14)| | | | ~(Fb & ~Rba) (15)| | | | Ga ~Rba (16)| | | | Fb ~~Rba Pr UD CD As ID As DD 6,~ ƎO 1, ƎO 9,&O 9,&O 11,~ ƎO 12, ∀O 8, ∀O 13,~&O 16,~&O Now, I am able to do an arrow-out using lines 4 and 15. I can also do an arrow-out using lines 10 and 16. I do these two steps now. (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) Pr UD CD As ID As DD 6,~ ƎO 1, ƎO (10)| | | | Fb (11)| | | | ~ Ǝ y(Gy & Rby) (12)| | | | ∀y~(Gy & Rby) (13)| | | | ~(Ga & Rba) (14)| | | | ~(Fb & ~Rba) (15)| | | | Ga~Rba (16)| | | | Fb~~Rba (17)| | | | ~Rba (18)| | | | ~~Rba 9,&O 9,&O 11,~ ƎO 12, ∀O 8, ∀O 13,~&O 16,~&O 4,15, O 10,16, O I now have a formula and its negation on lines 17 and 18 respectively. So I have found an X and need only then to complete my proof by crossing through my SHOWs. (1) Ǝ x[Fx & ~ Ǝ y(Gy & Rxy)] (2) SHOW: ∀x[Gx Ǝy(Fy & ~Ryx)] (3) | SHOW: Ga Ǝy(Fy & ~Rya) (4) | | Ga (5) | | SHOW: Ǝy(Fy & ~Rya) (6) | | | ~ Ǝy(Fy & ~Rya) (7) | | | SHOW: X (8) | | | | ∀y~(Fy & ~Rya) (9) | | | | Fb & ~ Ǝ y(Gy & Rby) (10)| | | | Fb (11)| | | | ~ Ǝ y(Gy & Rby) (12)| | | | ∀y~(Gy & Rby) (13)| | | | ~(Ga & Rba) (14)| | | | ~(Fb & ~Rba) (15)| | | | Ga~Rba (16)| | | | Fb~~Rba (17)| | | | ~Rba (18)| | | | ~~Rba (19)| | | | X Pr UD CD As ID As DD 6,~ ƎO 1, ƎO 9,&O 9,&O 11,~ ƎO 12, ∀O 8, ∀O 13,~&O 16,~&O 4,15,O 10,16,O 17,18,xI ...
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