Phi 270 F09 test 4

Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. Also restate your analyses using unrestricted quantifiers.

1.

Everyone saw the eclipse.

answer
2.

Al didn’t find any book that he was looking for.

answer
3.

No one ate only potato chips.

answer

Synthesize an English sentence that has the following logical form; that is, devise a sentence that would have the following analysis:

4.

(∀x: ¬ Sbx) Sax

S: [ _ saw _ ]; a: Al; b: Bill

answer

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

5.
∀x (Gx → Hx)
∀x (Fx ∧ Gx)
∀x Hx
answer
6.
∀y ∀x (Px → ¬ Fxy)
∀x ∀y (Fyx → ¬ Py)
answer

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

7.
∀x Rxa
∀x Rxx
answer

Phi 270 F09 test 4 answers

1.

everyone saw the eclipse

everyone is such that (he or she saw the eclipse)

(∀x: x is a person) x saw the eclipse

(∀x: Px) Sxe
∀x (Px→ Sxe)
P: [ _ is a person]; S: [ _ saw _ ]; e: the eclipse
2.

Al didn’t find any book that he was looking for

every book that Al was looking for is such that (he didn’t find it)

(∀x: x is a book that Al was looking for) Al didn’t find x

(∀x: x is a bookAl was looking for x) ¬ Al found x

(∀x: Bx ∧ Lax) ¬ Fax
∀x ((Bx ∧ Lax) → ¬ Fax)
B: [ _ is a book]; F: [ _ found _ ]; L: [ _ was looking for _ ]; a: Al
3.

no one ate only potato chips

no one is such that (he or she ate only potato chips)

(∀x: x is a person) ¬ x ate only potato chips

(∀x: Px) ¬ only potato chips are such that (x ate them)

(∀x: Px) ¬ (∀y: ¬ y is a potato chip) ¬ x ate y

(∀x: Px) ¬ (∀y: ¬ Cy) ¬ Axy
∀x (Px → ¬ ∀y (¬ Cy → ¬ Axy))
A: [ _ ate _ ]; C: [ _ is a potato chip]; P: [ _ is a person]
4.

(∀x: ¬ Bill saw x) Al saw x

(∀x: Bill didn’t see x) Al saw x

everything that Bill didn’t see is such that (Al saw it)

Al saw everything that Bill didn’t see

5.
│∀x (Gx → Hx)a:2
│∀x (Fx ∧ Gx)a:3
├─
│ⓐ
2 UI││Ga → Ha5
3 UI││Fa ∧ Ga4
4 Ext││Fa
4 Ext││Ga(5)
5 MPP││Ha(6)
││●
│├─
6 QED││Ha1
├─
1 UG│∀x Hx
6.
│∀y ∀x (Px → ¬ Fxy)a:5
├─
│ⓐ
││ⓑ
││││Fba(8)
│││├─
│││││Pb(7)
││││├─
5 UI│││││∀x (Px → ¬ Fxa)b:6
6 UI│││││Pb → ¬ Fba7
7 MPP│││││¬ Fba(8)
│││││●
││││├─
8 Nc│││││⊥4
│││├─
4 RAA││││¬ Pb3
││├─
3 CP│││Fba → ¬ Pb2
│├─
2 UG││∀y (Fya → ¬ Py)1
├─
1 UG│∀x ∀y (Fyx → ¬ Py)
7.
│∀x Rxaa:2, b:3
├─
│ⓑ
2 UI││Raa
3 UI││Rba
│││¬ Rbb
││├─
│││○¬ Rbb, Rba, Raa ⊭ ⊥
││├─
│││⊥4
│├─
4 IP││Rbb1
├─
1 UG│∀x Rxx

Counterexample presented by a diagram