## Description

Please complete all questions in problem set 9. You may also need problem set 8, lecture slides, and the textbook called Logic in Action. I will provide all of them in attachments.

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## Explanation & Answer

Please find the answer in the attachment below, (i was not able to the finish number 11 since i couldn't understand how it connects with the notes- i hope you understand, there was barely enough time to finish everything credibly). Anyway of you have any questions and clarrificarions please feel free to reach out

Surname 1

Name of student

Name of tutor

Course

Date

Problem Set 9

1

Syntax and Semantics of Monadic Predicate Logic

1.1

Identity and Substitution

1.1.1 The Identity Predicate

1. 4 points Exercise 4.36 of Logic in Action. Write out your formula with no

abbreviations.

∃x∃y∃z (¬x = y ∧ ¬x = z ∧ ¬x = y ∧ ∀v (P v ↔ (v = y ∨ v =

x ∨ v = z)))

2. 4 points Exercise 4.38(2) of Logic in Action. Demonstrate the diﬀerence in

meaning by providing a model in which the two formulas have diﬀerent

truth values.

In a model with two elements one of which is A but not B and

the other is B but not A the formulas have a different truth

value: ∃!x(Ax ∨ Bx) is false and ∃!x Ax ∨ ∃!x Bx is true.

Surname 2

1.1.2 Substitution

3. 10 points total; 2.5 points each part Give a recursive definition of the set of

terms t that are substitutable for a variable x in a formula ϕ. That is, fill in

the question marks in the following template (note that we do not assume

that the metavariables ‘x’ and ‘y’ refer to distinct variables, so you need to

analyze the case where x = y and the case where x ≠ y in part (d)):

t is always substitutable for x in an atomic formula P(u) or s

˙= t (because there are no quantifiers in an atomic formula to

capture a variable in t);

Solution

For any terms s and t and variable x, S xt recursive as follows:

Consider the term Biomom (x) which can be substituted to get

biomom (biomom (Alex)

If c ∈ Const, then cxt = c;

if y ∈ Var and y ≠ x, then yxt = y;

Xxt = t;

if f ∈ Func and u ∈ Term, then f (u) xt = f (u xt ).

t is substitutable for x in ¬ϕ iff ?

P(s) xt = P(sxt ), where Sxt is defined as before

For # ∈ {∧, ∨, →, ↔}, t is substitutable for x in (ϕ # ψ) iff?

(ϕ) xt = ϕxt and (ϕ # ꝕ) xt = (ϕ xt # ꝕ xt)

For Q ∈ {∀, ∃}, t is substitutable for x in Qyϕ iff?

(∀yP(x) ∧ ∃xQ(x)) xalex = (∀yP(x)) xalex ∧ (∃xQ(x)) xalex

2

Syntax and Semantics of Predicate Logic

4. These problems concern translating English sentences into the language of

predicate logic:

(a) 3 points assume the domain of discourse is all dogs (d). Translate the

following sentences into predicate logic: (Use f for Fido, r for Rover,

and L for loves)

Fido is loved by everyone.

∃x →Lxf

Rover loves everyone who

love Fido.

∀x (Rdf→LRd)

Fido and Rover love each

other.

Lfr ∧ Lrf

Surname 3

(b) 5 points for each of the following, specify an appropriate domain of

discourse, specify a translation key, and translate into predicate logic.

(Note: you do not have to understand what a sentence means before you

can attempt to translate it.)

Cats that love dogs don’t

taunt dogs.

∀x ((Cx → Lyx) ¬ ∀y(Ty)

There is a greatest natural

number.

∃x∀yG xy

Friends of Fido’s friends are

Rover’s friends.

∀xy ((Ffx ∧ Fxr) →Ffr)

There is no largest even number.

∀x(Ex→∃y(Ey ∧ Sxy ))

Every apple is tastier than every orange.

∀xy(Ax >Oy)

(c) 4 points translate the following sentences into predicate logical

formulas. Assume the domain of discourse is cats and dogs.

Some cat doesn’t love all dogs.

∃X ∃y (Cx ∧ Dy ∧ ¬ Lxy)

Every dog who loves all cats doesn’t

love every dog.

∀x((C x ∧ ∃y(Dy ∧ Lxy)) → ∃z(Dz ∧ Lzx))

Every cat who loves a dog is purred at

by some cat.

∀x((C x ∧ ∀y (Dv→Lxy)) → ∃z(Cz ∧ Lz...