Mathematical Proof and the Principles of Mathematics/Logic/The existential quantifier

If the universal quantifier is an extended version of conjunction, then what about an extended version of disjunction? This is the only other connective for which this type of extension makes sense, as you should be able to convince yourself. The existential quantifier We define the existential quantifier in terms of the universal quantifier by taking For some , to mean Not (for all , not ). The second statement is a bit complicated so lit's unpack it to see if the definition makes sense. For the statement For all , to be False there would be need to a value of for which is False. Applying this when is not , this says that For the statement For all , not to be False there would be need to a value of for which is True. In other words Not (for all , not ) is the same as saying there is a value of for which is True, or more briefly For some , There is another thing to notice about the statement Not (for all , not ) namely that it depends on the order of the quantifier and the two negations. More specifically, Not not (for all , ) and For all , not not are both equivalent to For all , which is not what we want. One way to interpret the existential quantifier logically is to say that For all x, P(x) is the same as the disjunction of all the statements P(a), where a takes on every value in the universe of discourse. If the universe is discourse is finite then this can be written out explicitly. So if the universe of discourse consists of two objects 'a' and 'b', then For some , is the same as or This makes sense in terms of our definition since or is equivalent to not (not and not ) If our universe of discourse has no objects, then the definition says that For some , is False no matter what is. This may seem a bit odd, but it does make sense since if you rephrase the statement as There is some for which then it would have to be False if the There is some part were False. If our universe of discourse has one object, then the definition says that For some , is the same as Not (for all , not ) or Not not or Coincidentally, it turns out that For some , is equivalent to For all , in a universe with one object. Proving an existential quantifier As usual with this type of definition, we get two rules of inference: From Not (for all , not ) deduce For some , ) and From For some , deduce Not (for all , not ) These are very practical for doing proofs though so we'll add some others based on these. First, we'll prove, as an example: Prop. 1: implies for some , Set up the proof of an implication in the usual way to get LineStatementJustification1 Hypothesis (something)n For some , ?n+1 implies for some , From 1 and n At this point we don't have anything but the definition to prove For some , so use that for the previous line. LineStatementJustification1 Hypothesis (something)n−1 Not (for all , not )?n For some , From n−1n+1 implies for some , From 1 and n Now we need to prove a negation, so assume the statement is true and derive a contradiction. LineStatementJustification1 Hypothesis2 For all , not Hypothesis (something)n−2 ?n−1 Not (for all , not )From 2 and n−2n For some , From n−1n+1 implies for some , From 1 and n But from 2 we can deduce not , which leads to the needed contradiction and the proof can be completed. LineStatementJustification1 Hypothesis2 For all , not Hypothesis3 not From 24 From 1 and 35 Not (for all , not )From 2 and 46 For some , From 57 implies for some , From 1 and 6 Combining Prop. 1 with the other rules of inference we get From derive For some , Using an existential quantifier The rule for using an existential quantifier is relatively complex. It's based on the following proposition which will be our second example proof. Prop. 2: (For some , ) and (for all , ( implies )) implies First set up the proof of an implication in the normal way, and break up the hypothesis in separate statements. Set up the proof of an implication in the usual way to get LineStatementJustification1 (For some , ) and (for all , ( implies ))Hypothesis2 For some , From 13 For all , ( implies )From 1 (something)n ?n+1(For some , ) and (for all , ( implies )) implies From 1 and n Use the definition to expand line 2 since that's the only way to use an existential so far. LineStatementJustification1 (For some , ) and (for all , ( implies ))Hypothesis2 For some , From 13 For all , ( implies )From 14 Not (for all , not )From 1 (something)n ?n+1(For some , ) and (for all , ( implies )) implies From 1 and n Now we need to prove , and since there doesn't seem to be a direct proof, so set up an indirect proof. LineStatementJustification1 (For some , ) and (for all , ( implies ))Hypothesis2 For some , From 13 For all , ( implies )From 14 Not (for all , not )From 15 Not Hypothesis (something)n−1 ?n From 5 and n−1n+1(For some , ) and (for all , ( implies )) implies From 1 and n In order to get to a contradiction, we prove that line 4 is False, in other words For all , not is True. This is a universal so introduce an arbitrary constant and derive not LineStatementJustification1 (For some , ) and (for all , ( implies ))Hypothesis2 For some , From 13 For all , ( implies )From 14 Not (for all , not )From 15 Not Hypothesis6 Choose Arbitrary constant (something)n−3 not ?n−2 for all , not From 6 and n−3n−1 From 4 and n−2n From 5 and n−1n+1(For some , ) and (for all , ( implies )) implies From 1 and n At this point can be plugged into line 3 and the rule of inference for statements can be used to complete the proof. LineStatementJustification1 (For some , ) and (for all , ( implies ))Hypothesis2 For some , From 13 For all , ( implies )From 14 Not (for all , not )From 15 Not Hypothesis6 Choose Arbitrary constant7 implies From 38 Not From 5 and 79 For all , not From 6 and 810 From 4 and 911 From 5 and 1012(For some , ) and (for all , ( implies )) implies From 1 and 11 In prose format: Prop. 2: (For some , ) and (for all , ( implies )) implies Proof: Assume for some , Also assume for all , ( implies We prove by contradiction so assume not To get a contradiction, it is enough to prove for all , not , since it the negation of the first assumption. Choose From the second assumption, implies But this and not imply not This holds for any , so for all , not and this, with the first assumption, is the required contradiction. By combining this proposition with the other rules of inference, including the rule for proving a universal, we can recast it as a rule of inference: If one has For some , P(x) and if by choosing as an arbitrary constant one can derive implies then deduce . In order to prove the implication, one would then assume and derive . It's easier to combine the Choose with the Assume steps into one to get Choose Read this as: Choose such that With this bit of streamlining, the rule of inference becomes If one has For some , and if by choosing one can derive , then deduce . For each and for every The two quantifiers can be combined it two ways: For some , for all , For all , for some , and For clarity, it is better to rephrase the first as There is some so that for every , For each , there is some so that and the second as Despite just being a change in the order of the quantifiers, these two statements are different. To see this, let the universe of discourse consist of magical rings and let stand for " rules ". The second statement merely states that every ring is ruled by some ring. One could easily imagine that every ring rules itself, so this wouldn't be saying much. But the first statement says there is a ring, which we might call the One Ring, which rules all the other rings; that's quite a ring. Note that we're using the word 'every' in the first statement to emphasis that the works for every , while in the second statement we're using the word 'each' to emphasize that the value of depends on So the second statement does not imply the first, but we can proof, as our third example, that the first statement implies the second. Prop. 3: There is some so that for every , implies for each , there is some so that Set up a direct proof in the usual way, then use the existential according to the new rule. LineStatementJustification1 There is some so that for every , Hypothesis2 Choose : for all , Constant satisfying condition (something)n−1 For each , there is some so that ?n For each , there is some so that From 1, 2 and n−1n+1There is some so that for every , implies for each , there is some so that From 1 and n We now need to prove For each , there is some so that which is a universal, so prove it for an arbitrary LineStatementJustification1 There is some so that for every , Hypothesis2 Choose : for all , Constant satisfying condition3 Choose Arbitrary constant (something)n−2 There is some so that ?n−1 For each , there is some so that From 3 and n−2n For each , there is some so that From 1, 2 and n−1n+1There is some so that for every , implies for each , there is some so that From 1 and n Now we can plug into line 2 and the proof is done. LineStatementJustification1 There is some so that for every , Hypothesis2 Choose : for all , Constant satisfying condition3 Choose Arbitrary constant 4 From 25 There is some so that From 46 For each , there is some so that From 3 and 57 For each , there is some so that From 1, 2 and 68There is some so that for every , implies for each , there is some so that From 1 and 7 In the prose version there's no need to repeat lines 6 and 7, so it becomes: Prop. 3: There is some so that for every , implies for each , there is some so that Proof: Assume there is some so that for every , Choose so that for every , and let be arbitrary. Then and so for some , This holds for arbitrary , therefore for each there is so that