Predicates and Quantifiers

Lecture 8, DSCI 220, 2026W1

Announcements

• EX1 is open — Thu 9/24 through Sun 9/27, self-scheduled in ORCA.
• Office hours are now Tue 10:30–11:30, ICCS 233.
• HW2 is released Sunday, due Sun Oct 4.

Three Endings

Contradiction

1	\(D\)
2	\(D \rightarrow \neg C\)
3	\(\neg C \wedge \neg B \rightarrow \neg E\)
4	\(\neg B\)
5	\(\neg G \vee \neg F\)
\(\therefore\)	\(\neg E\)

	Clause
1	\(\{D\}\)
2	\(\{\neg C,\ \neg D\}\)
3	\(\{B,\ C,\ \neg E\}\)
4	\(\{\neg B\}\)
5	\(\{\neg F,\ \neg G\}\)
6	\(\{E\}\)

Worksheet set 1.

D ⇒ -C ⇒ -B, then -E empties. The conclusion FOLLOWS – denying it was impossible.

Runs Out

1	\(\neg G\)
2	\(\neg G \rightarrow A\)
3	\(A \rightarrow \neg B\)
4	\(D \rightarrow C\)
\(\therefore\)	\(D\)

	Clause
1	\(\{\neg G\}\)
2	\(\{A,\ G\}\)
3	\(\{\neg A,\ \neg B\}\)
4	\(\{C,\ \neg D\}\)
5	\(\{\neg D\}\)

Worksheet set 6.

-G ⇒ A ⇒ -B ⇒ -D, and the last clause is satisfied. Nothing left, no contradiction. So the conclusion does NOT follow, and those four units are a counterexample: they satisfy every premise while the conclusion is false.

Stalls

1	\(\neg A\)
2	\(\neg A \rightarrow C\)
3	\(\neg B \vee \neg E\)
4	\(F \vee \neg G\)
5	\(\neg B \wedge F \rightarrow \neg D\)
\(\therefore\)	\(\neg D\)

	Clause
1	\(\{\neg A\}\)
2	\(\{A,\ C\}\)
3	\(\{\neg B,\ \neg E\}\)
4	\(\{F,\ \neg G\}\)
5	\(\{B,\ \neg D,\ \neg F\}\)
6	\(\{D\}\)

Worksheet set 10.

-A ⇒ C ⇒ D, then {-B,-E} {-G,F} {-F,B} remain and none is a unit. Propagation is done and has settled nothing. The conclusion does not follow, but we have not shown that – we have only run out of moves.

Predicates

\(p\): ___________ ate cereal for breakfast.

We may want to apply the statement to many students, so we define a predicate.

• \(P(x)\): \(x\) ate cereal for breakfast.

• \(x\) can be instantiated to a particular student, or an arbitrary one.

• \(P(\underline{\hspace{2em}})\) is a proposition.

The blank is the whole idea. A predicate is a proposition with a hole in it, and nothing is true or false until the hole is filled.

Go back to row 1 with this: P(x) = x is a kitten, C(x) = x is a cat. Now the sentence has a shape – but it still is not a proposition, because x is still loose. That is the next section.

Cereal

Boolean Masks

Select all the cereals with at least 4 units of protein and no more than 6 units of sugars.

Answer: cereals[~p & ~q], or equivalently ~(p | q) – De Morgan, and worth running both to show they agree.

THE POINT OF THE SLIDE. By filtering, they are defining a predicate instantiated by each row and evaluating its truth value. Say that sentence.

p and q are not propositions here. “A cereal has less than 4 units of protein” is not true or false until you say WHICH cereal – and the frame says which, 77 times over. Each row fills the blank, and pandas hands back the column of answers. That column IS the predicate, evaluated.

In week 3 we asserted ~(p v q) for Cheerios without noticing the problem: there are four Cheerios in this file, and it holds for only one of them. Plain Cheerios is protein 6, sugars 1. Apple Cinnamon is 2 and 10, Honey Nut 3 and 10, Multi-Grain 2 and 6. Worth filtering for “Cheerio” in the name if someone wants to see it.

Nine of the 77 survive the filter.

Quantifiers

Quantified Predicates

Consider the predicate

       \(H(x)\): \(x\) is a hero

Is this a proposition? ______

We add a quantifier to bind the variable:

  \(\forall\)   “for every”, “for all”
  \(\exists\)   “there exists”, “there is at least one”

       \(\forall x\, H(x)\)    \(\exists x\, H(x)\)

Now? ______

No, then yes. The variable is free in the first and bound in the second, and binding is what closes the hole.

Worth saying: a quantifier needs a domain. “Everyone is a hero” – everyone in what? For the cereal frame the domain was the 77 rows, which is why we could just count.

Counterexample

Someone says “all comedians are funny,” and you disagree.

       \(\forall x,\ C(x) \rightarrow F(x)\)

Refute by negating:

       \(\neg\forall x,\ C(x) \rightarrow F(x)\)
     \(\equiv \exists x,\ \neg(C(x) \rightarrow F(x))\)
     \(\equiv \exists x,\ \neg(F(x) \vee \neg C(x))\)
     \(\equiv \exists x,\ \neg F(x) \wedge C(x)\)

Your immediate response: ______________________

This evidence of refutation is called a _______________.

“No – what about THIS one?” You name a comedian who is not funny.

Counterexample. And note what it takes: ONE. To refute a “for all” you need a single case, which is why the negation of a universal is an existential:

~forall x P(x) == exists x ~P(x)

They met this shape in week 2 without knowing it – refuting “if you eat breakfast you ace the quiz” took exactly one morning where you did and did not.

Witness

Someone says “there is a funny comedian,” and you disagree.

       \(\exists x,\ C(x) \wedge F(x)\)

Their immediate response: ______________________

This evidence of justification is called a _______________.

Aside: why not \(\exists x,\ C(x) \rightarrow F(x)\)?

They name one. A witness.

So the asymmetry: one case SETTLES an existential and one case DESTROYS a universal. Neither can be settled the other way round without checking the whole domain – unless the domain is small enough to exhaust, which is exactly the distinction from lecture 1.

The aside: C(x) -> F(x) is true whenever C(x) is FALSE, so “there exists x with C(x) -> F(x)” is satisfied by any non-comedian. It claims almost nothing. With “exists” you want AND; with “for all” you want ->. That pairing catches everyone once.

Words and Ideas

Predicate, instantiation

Free and bound variables

Domain

\(\forall\), \(\exists\)

Counterexample, witness