Phi 270 F03 test 4 in pdf format
Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. Restate 2 using an unrestricted quantifier.
1. No one called the new number
[answer]
2. Sam asked everyone he could think of [Remember to restate this one using an unrestricted quantifier.]
[answer]
3. If any door was opened, the alarm sounded
[answer]
4. Only people who'd read everything the author had written were asked to review the book
[answer]
Use derivations to show that the following arguments are valid. You may use any rules.
5.
∀x (Fx ∧ Gx)
∀x Gx
[answer]
6.
(∀x: Fx) Gx
∀x ∀y (Gy → Rxy)
∀x (∀y: Fy) Rxy
[answer]
Use a derivation to show that the following argument is not valid and describe a structure (by using either a diagram or tables) that divides an open gap.
7.
(∀x: Fx) Rxa
Fa → ∀x Rxx
[answer]
Phi 270 F03 test 4 answers
1.

No one called the new number

No one is such that (he or she called the new number)

(∀x: x is a person) ¬ x called the new number)

(∀x: Px) ¬ Cxn
[C: λxy (x called y); P: λx (x is a person); n: the new number]
2.

Sam asked everyone he could think of

everyone Sam could think of is such that (Sam asked him or her)

(∀x: x is a person Sam could think of) Sam asked x

(∀x: x is a person ∧ Sam could think of x) Asx

( ∀x: Px ∧ Tsx) Asx
∀x ((Px ∧ Tsx) → Asx)
[A: λxy (x asked y); P: λx (x is a person); T: λxy (x could think of y); s: Sam]
3.

If any door was opened, the alarm sounded

every door is such that (if it was opened, the alarm sounded)

(∀x: x is a door) if x was opened, the alarm sounded

(∀x: Dx) (x was opened → the alarm sounded)

( ∀x: Dx) (Ox → Sa)
[D: λx (x is a door); O: λx (x was opened); S: λx (x sounded); a: the alarm]
4.

Only people who'd read everything the author had written were asked to review the book

Only people who'd read everything the author had written are such that (they were asked to review the book)

(∀x: ¬ x is a person who'd read everything the author had written) ¬ x was asked to review the book

(∀x: ¬ (x is a person ∧ x had read everything the author had written)) ¬ Axb

(∀x: ¬ (x is a person ∧ everything the author had written is such that (x had read it))) ¬ Axb

(∀x: ¬ (Px ∧ (∀y: y is a thing the author had written) x had read y)) ¬ Axb

(∀x: ¬ (Px ∧ (∀y: the author had written y) Rxy)) ¬ Axb

(∀x: ¬ (Px ∧ (∀y: Way) Rxy)) ¬ Axb
[A: λxy (x was asked to review y); P: λx (x is a person); R: λxy (x had read y); R: λxy (x had written y); a: the author; b: the book]
5.
│∀x (Fx ∧ Gx) a: 2
├─
│ⓐ
2 UI ││Fa ∧ Ga 3
3 Ext ││Fa
3 Ext ││Ga (4)
││ ●
│├─
4 QED ││Ga 1
├─
1 UG │∀x Gx
6.
│(∀x: Fx) Gx b:3
│ ∀x ∀y (Gy → Rxy) a:4
├─
│ⓐ
││ⓑ
│││Fb (3)
││├─
3 SB │││Gb (6)
4 UI │││∀y (Gy → Ray) b:5
5 UI │││Gb → Rab 6
6 MPP │││Rab (7)
│││●
││├─
7 QED │││Rab 2
│├─
2 RUG ││(∀y: Fy) Ray 1
├─
1 UG │∀x (∀y: Fy) Rxy
7.
│(∀x: Fx) Rxa a:2, b:5
├─
││Fa (2)
│├─
2 SB ││Raa
││ⓑ
││││¬ Rbb
│││├─
││││││¬ Fb
│││││├─
││││││○ Fa,Raa,¬Rbb,¬Fb ⇏ ⊥
│││││├─
││││││⊥ 6
││││├─
6 IP │││││Fb 5
││││
│││││Rba
││││├─
│││││○ Fa,Raa,¬Rbb,Rba ⇏ ⊥
││││├─
│││││⊥ 5
│││├─
5 MCR ││││⊥ 4
││├─
4 IP │││Rbb 3
│├─
3 UG ││∀x Rxx 1
├─
1 CP │Fa → ∀x Rxx
Counterexample presented by tables Counterexample presented by a diagram
range: 1, 2  
ab
12
 
τ Fτ 
1 T 
2 F 
 
R 1 2 
1 T F 
2 T F 

(This interpretation divides both gaps; the value of F2 is needed only for the 1st and the value of R21 only for the 2nd.)