phi_270_F02_test

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`]

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

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`

	│∀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

	│∀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)

	│∀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

a	b	c
1	2	3

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