Phi 270 F08 test 4

Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. State your analysis also in a form that expresses any generalizations using unrestricted quantifiers.

1.

No cover fit the container.

answer
2.

Everyone who Sam spoke to had seen the movie.

answer
3.

Only dogs chewed every bone.

answer
4.

No one who everyone knew bought anything.

answer

Use derivations to show that the following arguments are valid. You may use any rules.

5.
∀x (Fx → Hx)
∀x ((Fx ∧ Gx) → Hx)
answer
6.
∀x (Px → ∀y (Rxy → Txy))
∀x ∀y ((Px → Rxy) → (Px → Txy))
answer

Use a derivation to show that the following argument is not valid and present a counterexample by using a diagram to describe a structure that divides an open gap.

7.
∀x Rax
∀x (Rxx → Rxa)
answer

Phi 270 F08 test 4 answers
1.

no cover fit the container

no cover is such that (it fit the container)

(∀x: x is a cover) ¬ x fit the container

(∀x: Cx) ¬ Fxc
∀x (Cx→ ¬ Fxc)
C: [ _ is a cover]; F: [ _ fit _ ]; c: the container
2.

everyone who Sam spoke to had seen the movie

everyone who Sam spoke to is such that (he or she had seen the movie)

(∀x: x is a person who Sam spoke to) x had seen the movie

(∀x: x is a personSam spoke to x)) Sxm

(∀x: Px ∧ Ksx) Sxm
∀x ((Px ∧ Ksx) → Sxm)
K: [ _ spoke to _ ]; P: [ _ is a person]; S: [ _ had seen _ ]; m: the movie; s: Sam
3.

only dogs chewed every bone

only dogs are such that (they chewed every bone)

(∀x: ¬ x is a dog) ¬ x chewed every bone

(∀x: ¬ Dx) ¬ every bone is such that (x chewed it)

(∀x: ¬ Dx) ¬ (∀y: y is a bone) x chewed y

(∀x: ¬ Dx) ¬ (∀y: By) Cxy
∀x (¬ Dx → ¬ ∀y (By → Cxy))
B: [ _ is a bone]; C: [ _ chewed _ ]; D: [ _ is a dog]
4. No one who everyone knew bought anything

everything is such that (no one who everyone knew bought it)

∀x no one who everyone knew bought x

∀x no one who everyone knew is such that (he or she bought x)

∀x (∀y: y is a person who everyone knew) ¬ y bought x

∀x (∀y: y is a personeveryone knew y) ¬ Byx

∀x (∀y: Py ∧ everyone is such that (he or she knew y)) ¬ Byx

∀x (∀y: Py ∧ (∀z: z is a person) z knew y) ¬ Byx

∀x (∀y: Py ∧ (∀z: Pz) Kzy) ¬ Byx
∀x ∀y ((Py ∧ ∀z (Pz → Kzy)) → ¬ Byx)
B: [ _ bought _ ]; K: [ _ knew _ ]; P: [ _ is person]
5.
│∀x (Fx → Hx) a:4
├─
│ⓐ
│││Fa ∧ Ga 3
││├─
3 Ext │││Fa (5)
3 Ext │││Ga
4 UI │││Fa → Ha 5
5 MPP │││Ha (6)
│││●
││├─
6 QED │││Ha 2
│├─
2 CP ││(Fa ∧ Ga) → Ha 1
├─
1 UG │∀x ((Fx ∧ Gx) → Hx)
6.
│∀x (Px → ∀y (Rxy → Txy)) a:6
├─
│ⓐ
││ⓑ
││││Pa → Rab 5
│││├─
│││││Pa (5), (7)
││││├─
5 MPP │││││Rab (9)
6 UI │││││Pa → ∀y (Ray → Tay) 7
7 MPP │││││∀y (Ray → Tay) b:8
8 UI │││││Rab → Tab 9
9 MPP │││││Tab (10)
│││││●
││││├─
10 QED │││││Tab 4
│││├─
4 CP ││││Pa → Tab 3
││├─
3 CP │││(Pa → Rab) → (Pa → Tab) 2
│├─
2 UG ││∀y ((Pa → Ray) → (Pa → Tay)) 1
├─
1 UG │∀x ∀y ((Px → Rxy) → (Px → Txy))
7.
│∀x Rax a:3, b:4
├─
│ⓑ
│││Rbb
││├─
3 UI │││Raa
4 UI │││Rab
│││
││││¬ Rba
│││├─
││││○ ¬ Rba, Rab, Raa, Rbb ⇏ ⊥
│││├─
││││⊥ 5
││├─
5 IP │││Rba 2
│├─
2 CP ││Rbb → Rba 1
├─
1 UG │∀x (Rxx → Rxa)

Counterexample presented by a diagram