Phi 270 F02 test 4
Analyze the sentences below in as much detail as possible, providing a key to the non-logical vocabulary you use. Notice the special instructions for 2.
1. Only bears performed.
answer
2. If everyone cheered, the elephant bowed. [In this case, restate your answer using an unrestricted quantifier.]
answer
3. No one laughed at any performers except clowns.
answer
Synthesize an English sentence with the following logical form:
4. (∀x: Px ∧ Cxt) Ctx
C: [ _ called _ ]; P: [ _ is a person]; t: Tom
answer
Use derivations to establish the validity of the following arguments. You may use attachment rules.
5.
∀x Fx
∀x ¬ (Fx ∧ Gx)
∀x ¬ Gx
answer
6.
∀x ∀y (Fy → Rxy)
∀x (Fx → ∀y Ryx)
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 one of the derivation’s open gaps.
7.
∀x Rax
∀x (Rbx → ¬ Rxa)
∀x ¬ Rbx
answer

Phi 270 F02 test 4 answers
1.

Only bears performed

(∀x: ¬ x is a bear) ¬ x performed

(∀x: ¬ Bx) ¬ Px
B: [ _ is a bear]; P: [ _ performed]
2.

If everyone cheered, the elephant bowed

everyone cheered → the elephant bowed

(∀x: x is a person) x cheered → the elephant bowed

(∀x: Px) Cx → Be
∀x (Px → Cx) → Be
B: x bowed; C: x cheered; P: x is a person; e: the elephant
Incorrect:
(∀x: Px) (Cx → Be) or: ∀x (Px → (Cx → Be))
these say: If anyone cheered, the elephant bowed
3.

No one laughed at any performers except clowns

all performers except clowns are such that (no one laughed at them)

(∀x: x is a performer ∧ ¬ x is a clown) no one laughed at x

(∀x: x is a performer ∧ ¬ x is a clown) (∀y: y is a person) ¬ y laughed at x

(∀x: Fx ∧ ¬ Cx) (∀y: Py) ¬ Lyx
C: [ _ is a clown]; F: [ _ is a peformer]; P: [ _ is a person]; L: [ _ laughed at _ ]
Incorrect:
(∀y: Py) ¬ (∀x: Fx ∧ ¬ Cx) Lyx
says: No one laughed at all performers who weren’t clowns
4.

(∀x: x is a person ∧ x called Tom) Tom called x

(∀x: x is a person who called Tom) Tom called x

everyone who called Tom is such that (Tom called him or her)

Tom called everyone who called him
5.
│∀x Fx a:2
│∀x ¬ (Fx ∧ Gx) a:3
├─
│ⓐ
2 UI ││Fa (4)
3 UI ││¬ (Fa ∧ Ga) 4
4 MPT ││¬ Ga (5)
││●
│├─
5 QED ││¬ Ga 1
├─
1 UG │∀x ¬ Gx
6.
│∀x ∀y (Fy → Rxy) b:4
├─
│ⓐ
│││Fa (6)
││├─
│││ⓑ
4 UI ││││∀y (Fy → Rby) a:5
5 UI ││││Fa → Rba 6
6 MPP ││││Rba (7)
││││●
│││├─
7 QED ││││Rba 3
││├─
3 UG │││∀y Rya 2
│├─
2 CP ││Fa → ∀y Rya 1
├─
1 UG │∀x (Fx → ∀y Ryx)
7.
│∀x Rax a:3,b:4,c:5
│∀x (Rbx → ¬ Rxa) c:6,a:8,b:10
├─
│ⓒ
│││Rbc (7)
││├─
3 UI │││Raa (9)
4 UI │││Rab
5 UI │││Rac
6 UI │││Rbc → ¬ Rca 7
7 MPP │││¬ Rca
8 UI │││Rba → ¬ Raa 9
9 MTT │││¬ Rba
10 UI │││Rbb → ¬ Rba 11
│││
│││││¬ Rbb
││││├─
│││││○ Rbc,Raa,Rab,Rac,¬Rca,¬Rba,¬Rbb ⇏ ⊥
││││├─
│││││⊥ 12
│││├─
12 IP ││││Rbb 11
│││
││││¬ Rba
│││├─
││││○ Rbc,Raa,Rab,Rac,¬Rca,¬Rba ⇏ ⊥
│││├─
││││⊥ 11
││├─
11 RC │││⊥ 2
│├─
2 RAA ││¬ Rbc 1
├─
1 UG │∀x ¬ Rbx
Counterexample presented by tables Counterexample presented by a diagram
range: 1, 2, 3  
abc
123
 
R 1 2 3
1 T T T
2 F F T
3 F F F
Grayed values are not required to divide either gap;
the value for R22 is not required to divide the 2nd gap