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Metalogic
Soundness and Completeness
Two Notions of Logical Consequence
Validity: If the premises are true, then the conclusion must be true.
Provability: From the premises, the conclusion can be derived by the
rules of inference of the system.
Two Notions of Logical Consequence
Symbol for validity: ⊨
Example: Q ⊨ (P → Q)
Symbol for provability: ⊢
Example: Q ⊢ (P → Q)
Two Questions about Equivalence
Soundness:
Is it always true that if φ ⊢ ψ, then φ ⊨ ψ ?
Completeness:
Is it always true that if φ ⊨ ψ, then φ ⊢ ψ ?
Soundness
Soundness
Soundness:
Is it always true that if φ ⊢ ψ, then φ ⊨ ψ ?
Why are we interested in soundness?
(What’s wrong with a system that proves false things from true
premises?)
Proving Soundness
It’s not possible to prove that all logical systems are sound. Sometimes
(as with arithmetic) we just assume they are.
But it’s nice to have a proof!
The Rule of Assumption: A
Assumption is the easiest rule to learn. It says at any stage in the
derivation, we may write down any WFF we want to. That WFF only
depends on itself. For example, on line 57 we might write:
57
57. (((P ↔ (A & B)) → ~~~R)
A
Turnstile
The proper way to read this is “line 57 is provable from line 57” or “On
the assumption that (((P ↔ (A & B)) → ~~~R) it can be proved that (((P
↔ (A & B)) → ~~~R).”
We can re-write any line to state preciesely what we have proved. Line
57 would be:
(((P↔(A & B))→~~~R) Ⱶ (((P↔(A & B))→~~~R)
Proving Assumption is Sound
A sound rule is one where the truth of the premises guarantees the
truth of the conclusion.
Is it always true that if φ ⊢ φ, then φ ⊨ φ ?
&-Elimination: &E
&E is also a very easy-to-learn rule.
If we have proved (φ & ψ) on some line,
then on any future line we may write down φ,
and on any future line we may write down ψ.
The result depends on everything (φ & ψ) depended on.
Moving Pieces
The &E rule introduces a new feature: dependencies.
To get the soundness proof correct, we have to track them.
&E
So what does this mean:
“If we have proved (φ & ψ) on some line…” ?
It means that some line of our proof looks like this:
χ1, χ2,… , χn ⊢ (φ & ψ)
&E
And this means?
“then on any future line we may write down φ,
and on any future line we may write down ψ.
The result depends on everything (φ & ψ) depended on.”
&E
It means that these are allowable future lines:
χ1, χ2,… , χn ⊢ φ
χ1, χ2,… , χn ⊢ ψ
&E
So here’s our rule again:
When we have proved:
χ1, χ2,… , χn ⊢ (φ & ψ)
Then we can prove:
χ1, χ2,… , χn ⊢ φ
χ1, χ2,… , χn ⊢ ψ
Mathematical Induction
In mathematical induction we prove that 0 has some property.
Then we prove that if every number before x has that property, then x
has that property too.
Then we conclude that all numbers have that property.
Mathematical Induction
When you begin a proof (on line 0), the only possible thing to write
down are your assumptions.
All the other rules require that you’ve already proved something.
But we know that line 0 is sound, because we’ve already shown
assumption is sound.
Mathematical Induction
We know that line 0 is sound.
We assume that all lines before x are sound.
We prove that x then must be sound as well.
We conclude that all lines are sound.
&E
So here’s our rule again:
When we have proved:
χ1, χ2,… , χn ⊢ (φ & ψ)
Then we can prove:
χ1, χ2,… , χn ⊢ φ
χ1, χ2,… , χn ⊢ ψ
Assume this line (and those
before it) are sound.
Prove that future lines like
these are sound.
&E
So here’s our rule again:
When we have proved:
χ1, χ2,… , χn ⊢ (φ & ψ)
Then we can prove:
χ1, χ2,… , χn ⊢ φ
χ1, χ2,… , χn ⊢ ψ
Assume that if χ1, χ2,… , χn
are all true, then (φ & ψ) is.
Prove that if χ1, χ2,… , χn
are all true, then φ is.
Soundness for &E
Prove that if χ1, χ2,… , χn are all true, then φ is.
OK, assume for conditional proof that χ1, χ2,… , χn are all true.
Now, since we are also assuming that if χ1, χ2,… , χn are all true, then
(φ & ψ) is.
Thus it follows that (φ & ψ) is true.
Conjunction
φ
T
T
F
F
ψ
T
F
T
F
(φ & ψ)
T
F
F
F
Finish the Induction
We have to do similar proofs for all the rules, because we have to show
on any future allowable line, it’s still sound if all the earlier ones were.
Circularity?
Soundness for &E
Prove that if χ1, χ2,… , χn are all true, then φ is.
OK, assume for conditional proof that χ1, χ2,… , χn are all true.
Now, since we are also assuming that if χ1, χ2,… , χn are all true, then
(φ & ψ) is.
Thus it follows that (φ & ψ) is true.
Soundness for &E
Prove that if χ1, χ2,… , χn are all true, then φ is.
OK, assume for conditional proof that χ1, χ2,… , χn are all true.
Now, since we are also assuming that if χ1, χ2,… , χn are all true, then
(φ & ψ) is.
Thus it follows that (φ & ψ) is true.
Circularity
In order to prove that our logic rules are correct, we seem to be using
those very same rules (albeit not formally stated, and not in a rigorous
formal system).
Can logic really justify itself?
Completeness
Completeness
Completeness:
Is it always true that if φ ⊨ ψ, then φ ⊢ ψ ?
Notice that this isn’t the same as saying every valid argument is
provable, because that’s impossible. (Why?)
System Relativity
Soundness and completeness are properties of particular logical
systems.
There’s no sense to be made of the claim that “logic” is sound or
complete. Some logics are sound, others are not, some are complete,
others are not.
Completeness
Why do we want completeness in a system?
We know we don’t need anymore rules!
Completeness
Two logical systems that Papineau looks at are complete: propositional
logic and predicate (1st order) logic.
But not every logical system is complete: in fact, Goedel showed that
second-order logic and arithmetic are incomplete.
(Mention Decidability)
Final Exam: Answers
Multiple Choice
T
F
It’s possible that there is a village where in that village,
there is a barber, and the barber shaves all the villagers
who don’t shave themselves.
1.
Answer: T. Explanation: As we went over in class, the barber paradox
only arises from the assumption that the barber shaves all and only
villagers who don’t shave themselves.
Class Score: 48%
Multiple Choice
T
F
The continuum hypothesis says that for any infinite set S,
the members of the power set of S cannot be paired oneto-one with the members of S.
2.
Answer: F. Explanation: The power set theorem says that for any set
(not just infinite ones), the members of P(S) can’t be paired one-to-one
with the members of S. The continuum hypothesis says there is no set
larger than the natural numbers but smaller than the reals (in the oneto-one pairing notion of size).
Class Score: 52%
Short Answer
3. Write names of the following set in extensive notation:
{1, {}} ∩ {}
ANSWER: {}. Explanation: The intersection of A and B contains all the
members in both A and B. Since no members are in the null set, the
intersection of any set with the null set is equal to the null set.
Class Score: 91%
Union & Intersection
4. {x: x is an even number}, {x: x is a prime number}
UNION: {x: x is an even number OR x is a prime number}
INTERSECTION: {2}
Class Score: 87% (union), 91% (intersection)
Union & Intersection
5. {x: x owns a house}, {x: x owns a car}
UNION: {x: x owns a house OR x owns a car}
INTERSECTION: {x: x owns a house AND x owns a car}
Class Score: 91% (union), 96% (intersection)
Union & Intersection
6. {x: x is a philosopher}, {}
UNION: {x: x is a philosopher}
INTERSECTION: {}
Class Score: 96% (union), 96% (intersection)
Short Answer
7. Let S = {x: x is a natural number & 0 < x ≤ 10}. How many members
are in the power set of S?
ANSWER: 210 = 1024
Class Score: 91%
Analytic or Synthetic
A
S
Nothing can be red and blue all over.
8.
Answer: Any. I don’t think any answer is clearly right or clearly wrong.
Sometimes, even on exams, I ask you questions solely to get you to
think about things.
Class Score: N/A
Analytic or Synthetic
A
S
If person A is person B’s male sibling, then A is B’s brother.
9.
Answer: A. Explanation: Not needed because…
Class Score: 100%
Analytic or Synthetic
A
S
Dogs have four legs.
10.
Answer: S. Explanation: If it were analytic that dogs have four legs, then
three-legged dogs would be contradictions. But my friend Bryan has a
three-legged dog. So it’s not analytic. See Paul Ziff 1960 Semantic
Analysis pp. 184-185 or Saul Kripke 1980 Naming and Necessity pp.
119-120 for a similar argument regarding 3-legged tigers.
Class Score: 83%
Necessary or Contingent
N
C
If there is a 70% probability of rain tomorrow, then there is
a 30% probability of no rain.
11.
Answer: N. A probability function is a function that satisfies the axioms
of probability. From these axioms you can prove that Pr(not-P) = 1 –
Pr(P). Thus in any world where there is a probability function that
assigns 70% to the chance of rain tomorrow, it will also be true that
that function assigns 30% to the chance of no rain.
Class Score: 96%
Necessary or Contingent
N
C
Whales breathe oxygen.
12.
Answer: C. It doesn’t seem plausible to say that you have to breathe
oxygen to be a whale. Suppose tomorrow a scientist discovers that one
species which we have thus far been calling a species of “whale”
actually breathes nitrogen. Suppose that the species is still very closely
related to other whales and there’s no way of drawing the whale family
tree that excludes this species and includes all the other whales. I’m
inclined to think this species would still be a species of whale.
Class Score: 74%
Necessary or Contingent
N
C
Water is on Earth, but not on Twin Earth.
13.
Answer: C. Take just the first part of the conjunction: water is on Earth.
Obviously this is contingent. Some scientists even think water was not
on the early Earth, but was transported here by asteroids. One student
asked if I meant, “the water-like stuff on Earth is water,” but that’s no
way of reading that sentence, and it’s also contingent. The water-like
stuff on Earth could have been XYZ, had things gone different.
Class Score: 74%
True or False (Green Means Correct)
T
F
T
F
T
F
T
F
If term t is a rigid designator, then it is a (de dicto) necessity:
t = t. Class Score: 52%
If term t is a rigid designator, then it is a (de re) necessity: t = t.
Class Score: 91%
If term t is not a rigid designator, then it is a (de dicto)
necessity: t = t. Class Score: 65%
If term t is not a rigid designator, then it is a (de re) necessity:
t = t. Class Score: 78%
Discussion
Consider the de re claim: (necessarily)(Michael = Michael). Since
‘Michael’ is a rigid designator, it picks out the same person (me) at
every possible world. Since that person is identical to himself at each
world, the claim Michael = Michael is necessariy.
Consider the de dicto claim: (the x: x = Michael)(Necessarily)( =
Michael). This claim is also true. It is an essential fact of me that I am
identical to Michael.
Discussion
Now take a non rigid designator, like ‘the instructor of Phil Tools in the
second term of the 2014-2015 school year at HKU’. (Let’s abbreviate
that as ‘the instructor’).
It’s true that (necessarily)(the instructor = the instructor). For each
possible world, the person who is the instructor at that world is the
instructor at that world.
Discussion
But the de dicto claim is not true:
(the instructor)(necessarily)( = the instructor)
This says that the guy who is actually the instructor– Michael–
essentially is the instructor. But that’s not an essential property of me. I
might have taught a different class or no class at all this term.
True or False
T
F
Kant thought that geometry was analytic a priori. 18.
Answer: F. Explanation: Kant thought that geometry was synthetic a
priori.
Class Score: 57%
True or False
T
F
If p is not possible, then not-p is necessary.
19.
Answer: T. If p is not possible, then there is no world where p is true.
Thus in every world, not-p is true. Therefore, not-p is necessary.
Class Score: 96%
True or False
T
F
Necessarily, if Owen and Constance Papineau have an
oldest son, then that son is David Papineau.
Answer: F. If D.P. exists at a world, then he is the son of Owen and
Constance, according to the essentiality of origins thesis. This doesn’t
mean that they necessarily had to have him as their first son. They
could have had a different son before giving birth to D.P. They could
have had a different son, and never had D.P.
Class Score: 70%
Multiple Choice
21. Which of the following principles is the Identity of Indiscernibles?
a. (For all things x, y)(if x = y, then (for all properties F)(x has F iff y has F))
b. (For all things x, y)(if (for all properties F)(x has F iff y has F), then x = y)
Answer: b. See Box 14 p. 84 of the course textbook.
Multiple Choice
22. Which of the following principles is uncontroversial?
a. (For all things x, y)(if x = y, then (for all properties F)(x has F iff y has F))
b. (For all things x, y)(if (for all properties F)(x has F iff y has F), then x = y)
Answer: a. See Box 14 p. 94 of the course textbook.
Probability Theory
Suppose Pr(p) = 0.8, Pr(q) = 0.4, and Pr(p and q) = 0.3.
23. What is Pr(not p) ?
Pr(not p) = 1 – Pr(p) = 1 – 0.8 = 0.2
Class Score: 100%
Probability Theory
Suppose Pr(p) = 0.8, Pr(q) = 0.4, and Pr(p and q) = 0.3.
24. What is Pr(p or q) ?
Pr(p or q) = Pr(p) + Pr(q) – Pr(p and q) = 0.8 + 0.4 – 0.3 = 0.9
Class Score: 96%
Probability Theory
Suppose Pr(p) = 0.8, Pr(q) = 0.4, and Pr(p and q) = 0.3.
25. What is Pr(p and not q) ?
Solution: next page
Probability Theory
It’s a logical truth that p if and only if ((p & q) or (p & not q)). Thus:
Pr(P) = Pr((p & q) or (p & not q)) = 0.8
Since (p & q) and (p & not q) are incompatible, by Axiom 3:
Pr(p & q) + Pr(p & not q) = 0.8
0.3 + Pr(p & not q) = 0.8
Pr(p & not q) = 0.8 – 0.3 = 0.5
Class Score: 78%
Probability Theory
26. Suppose Pr(p / q) = ½. Pr(p) = 1/6, and Pr(q) = ⅓. What is Pr(q / p) ?
Pr(p / q)
= Pr(q / p)Pr(p) ÷ Pr(q)
= [½ x ⅓] ÷ 1/6
= 1/6 ÷ 1/6
=1
Class Score: 78%
Bayes’ Theorem!
Probability Theory
27. Suppose Pr(p) = 0.8, Pr(q) = 0.6, and Pr(p and q) = 0.3.
What is Pr(p / q) ?
Pr(p / q)
= Pr(p & q) ÷ Pr(q)
= 0.3 ÷ 0.6
= 0.5
Class Score: 91%
Def. of Conditional Probability!
Probability Theory
28. Suppose that your subjective degree of belief in q is 1/5 and your
subjective degree of belief in (p & q) is 1/10. According to
conditionalization, if you learn that q is true, what should your new
subjective degree of belief in p be?
According to conditionalization, when you learn q, for any p:
PrNEW(p) = PrOLD(p / q)
Probability Theory
According to conditionalization, when you learn q, for any p:
PrNEW(p) = PrOLD(p / q)
PrOLD(p / q)
= PrOLD(p and q) ÷ PrOLD(q)
= 1/10 ÷ 1/5 = 1/2
Class Score: 74%
Def. of Cond. Prob.
Multiple Choice
29. Which of the following statements is a theorem of probability
theory (the Law of Total Probability)?
a.
b.
c.
d.
Pr(p) = Pr(p / q) x Pr(p) + Pr(~p / q) x Pr(~p)
Pr(p) = Pr(p / q) x Pr(q) + Pr(p / ~q) x Pr(~q)
Pr(p) = Pr(q / p) x Pr(p) + Pr(q / ~p) x Pr(~p)
Pr(p) = Pr(q / p) x Pr(q) + Pr(~q / p) x Pr(~q)
We did an entire problem set on this one! Class Score: 78%
Multiple Choice
30. If A is positively correlated with B, then which of the following must
be true?
a.
b.
c.
d.
B is positively correlated with A.
B is negatively correlated with A.
A and B are positively correlated with another variable C.
None of the above.
Proof on next page. Class Score: 70%
Correlation
Suppose Pr(A / B) > Pr(A)
Then Pr(A / B) ÷ Pr(A) > 1
But Pr(B / A) = [Pr(A / B) ÷ Pr(A)] x Pr(B)
Bayes’ Theorem
So Pr(B / A) > Pr(B)
(Because you have to multiply Pr(B) by a factor greater than 1 to get
Pr(B / A).)
Pr(A / B) > Pr(A) if and only if Pr(B / A) > Pr(B)
31. Suppose I notice that two students have turned in exams with
exactly the same answers on the multiple choice questions. Suppose
also that I know this:
• If the two students cheated, the probability that their answers would
be the same is 100%.
• If the two students did not cheat, then the probability that their
answers would be the same is 2.5%.
32. What is the probability that the two students cheated, given that
their answers are the same?
a.
100%.
b.
97.5%
c.
2.5%
d.
It is impossible to tell, given the information in the question.
Explanation
In the problem you’re given the probability of same answers, given
cheating Pr(S / C) and the probability of same answers, given not
cheating Pr(S / not C).
You are asked to solve for the probability of cheating, given same
answers Pr(C / S).
But this depends both on the two givens and the “base rate” or prior
probability of cheating Pr(C). You are not given that. Class Score: 96%
Multiple Choice
33. Which of the following is most likely to happen?
a.
b.
c.
Everyone will fail this class.
Everyone will fail this class, because the instructor becomes evil.
The HKU president demands that the instructor fails everyone in
this class, and so he does.
d.
There is not enough information to answer this question.
Class Score: 70%
Explanation
Pr(A & B) is always less than or equal to Pr(A). See box 16 p. 95 of the
course textbook. Furthermore, when A doesn’t entail A & B, Pr(A & B)
is always strictly less than Pr(A).
This is common sense. If answer (b) happens, answer (a) also happens,
because (b) is (a) and the instructor is evil. If (c) happens, then (a)
happens, because (c) is (a) and the president demands it.
Number of Students w/ each Mark
Score
3.5
3
2.5
2
1.5
1
0.5
0
less than 27
27
28
29
30
Score
31
32
33
34
Remarks
All but one student missed between 2 and 10 questions (scores
between 34 and 26 out of 36). That’s between 94% and 72%. That’s
pretty good. Probably, the test was a little on the easy side, though
there were only two questions that everyone got right.
On the whole, good job.