1. Which of the following three schemata is equivalent with which other(s)?

           p ⊃ (q ≡ r),   p.q ≡ p.r,  and   p ∨ q ≡ p ∨ r

2. Determine whether the following argument is truth-functionally valid:

   Jack will go to the party only if Jill does but Tony does not.

   Jill will go to the party only if Tony and Jack do.

   Conclusion: Neither Jack nor Jill will go to the party.

3. If the biconditional of R and S is satisfiable, then R and S are either both satisfiable or both unsatisfiable.
4. Paraphrase the following using the existential or the universal quantifier:
   1. There is a jazz musician who plays the harp.

      JÀ	:	À is a jazz musician
      HÀ	:	À plays the harp

   2. Everyone who went to the concert bought the record.

      CÀ	:	À went to the concert
      BÀ	:	À bought the record

1. Paraphrase the following twice, once as a negated existential quantification, once as a universal quantification:
   No one in the room knows either Malone or Sutherland.

   RÀ	:	À is in the room
   MÀ	:	À knows Malone
   SÀ	:	À knows Sutherland

2. Paraphrase the following using quantificational notation:
   1. Some who take logic and Latin take neither physics nor Greek, but all who take either Latin or chemistry take both logic and Greek.

      LÀ	:	À takes logic	PÀ	:	À takes physics
      TÀ	:	À takes Latin	GÀ	:	À takes Greek
      CÀ	:	À takes chemistry

   2. Any student who is not both clever and persistent will have difficulty with the course.

      SÀ	:	À is a student
      CÀ	:	À is clever
      PÀ	:	À is persistent
      DÀ	:	À will have difficulty with the course

   3. Amnesty International adopts a political prisoner just in case she has neither used nor advocated violence.

      IÀ	:	Amnesty International adopts À
      PÀ	:	À is a political prisoner
      UÀ	:	À has used violence
      AÀ	:	À has advocated violence

3. II.A.5(b), paraphrase, once with the universe of discourse restricted to persons, once with unrestricted universe of discourse:
   
   One who acts in haste will repent in sorrow.
    

   HÀ	:	À acts in haste
   RÀ	:	À will repent in sorrow
   PÀ	:	À is a person

4. Paraphrase the following statements, taking the universe of discourse to be the class of all persons, and using the given abbreviations of the English predicates:

   SÀ	:	À is a soprano
   TÀ	:	À is a tenor
   RÀÁ	:	À respects Á
   LÀÁ	:	À is louder than Á

   1. [III.A.2(a)] A soprano who respects all tenors doesn’t respect herself.
   2. [III.A.2(e)] Some sopranos respect only those tenors who are louder than they are.
5. Show that p ⊃ (q ⊃ r) and (p ⊃ q) ⊃ r are not equivalent.

1. With universe of discourse and predicate letter abbreviations as in the last exercise, translate the following statements of logical English into clear, idiomatic English:
   1. (∃x)(∃y)(Tx.Rxy.Sy) ⊃ (∃y)[Ty.(∀x)(Sx ⊃ Rxy)]
   2. (∀x)[Sx ⊃ (∀y)(Ty⊃ (Rxy ≡ −Lxy))]
2. Taking the universe of discourse to be the class of persons, and using

   “LÀÁ” for “À loves Á”,

   paraphrase:

   1. Someone everyone loves loves someone.
   2. Everyone someone loves loves everyone.
3. Taking the universe of discourse to be the class of persons, and using

   PÀÁ	:	À is a parent of Á
   MÀ	:	À is male
   IÀÁ	:	À is identical to Á (i.e., À is the same person as Á)

   paraphrase:

   1. x is y’s paternal grandmother
   2. x and y are cousins (caution: no one is her own cousin!)

   NB. What you are asked to paraphrase in this exercise are OPEN SENTENCES, i.e., sentences in which VARIABLES OCCUR FREE. Hence, correct paraphrases must have exactly the same free variables as the original open sentences.

4. For each of the following pairs of schemata, show that the first does not imply the second by providing a structure with universe of discourse {1, 2, 3} in which the first schema is true and second schema is false:
   1. [II.B.3(b)]  (∀x)(Fx.Gx ⊃ Hx)    (∃x)Fx.(∃x)Gx ⊃ (∃x)Hx
   2. [II.B.3(f)]  (∃x)(Fx ≡ Gx)             (∃x)Fx ≡ (∃x)Gx

1. II.B.4: Give an interpretation under which (∀x)(Fx ∨ Gx ⊃ Hx ∨ Jx) is ⊤, but each of (∀x)(Fx ⊃ Hx), (∀x)(Fx ⊃ Jx), (∀x)(Gx ⊃ Hx), and (∀x)(Gx ⊃ Jx) is ⊥.
2. Taking the universe of discourse to be { 1,2,3}, find, for each of the following schemata, an extension of HÀÁ” which makes it false.
   1. (∀x)[(∃y)Hxy ⊃ (∃y)Hyx]
   2. (∀x)[(∀y)(Hyx ⊃ Hxy) ⊃ (∀y)(Hxy ⊃ Hyx)]
   3. (∀x)[(∃y)Hxy ⊃ Hxx]
   4. (∃x)(∀y)[Hyy ⊃ Hxy]
3. [5 pts each] For each of the following pairs of schemata, give an interpretation which shows that the first does not imply the second.

   (i)	(∃x)(∀y)−Fxy . (∃x)(∀y)Fxy		(∀x)((∃y)Fxy ⊃ (∃ y)Fyx)
   (ii)	(∀y)(∀x)[Fyx ⊃ (∃z)Fzy]		(∀y)[(∀z)Fyz ⊃ (∀ z)Fzy]
   (iii)	(∀x)(∃y)(Gxy . −Gyx)		(∀x)[(∃y)Gxy ⊃ (∃y)Gyx]
   (iv)	(∀x)[−Lx ⊃ (∃y)(Ly . Ayx)]		(∀x)(−Lx ⊃ (∀y)(Ly ⊃ Ayx))

4. Is the second schema of each of the following pairs of a schemata an instance of the first? If so, what is the instantial variable, and is the instance conservative?

   (i)	(∀x)(Gyx . Hxz)		Gyy . Hyz
   (ii)	(∀x)[(∃z)Hxz . (∃y)Fyy]		(∃z)Hyz . (∃y)Fyy
   (iii)	(∀x)[(∃z)Hxz . (∃y)Fyx]		(∃z)Hzz . (∃y)Fyz
   (iv)	(∀x)Fxz ∨ (∃y)Gy		Fzz ∨ (∃y)Gy

Exercises on General Laws:

1. Prove the Law of Existential Implication: If a schema Φ(u) implies a schema S and S does not contain u free, then (∃u)Φ(u) implies S.
2. Using the Law of Existential Implication and the Re-Lettering Law, prove that if (i) a schema Φ(u) implies a schema S, (ii) S does not contain u free, and (iii) Φ(u) is a conservative instance of (∃v)Φ(v), then (∃v)Φ(v) implies S.

1. (∀x)(∀y)(Fxy ⊃ −Fyx) implies (∀x)−Fxx
2. (∀x)(∀y)(∀z)(Fxy.Fyz ⊃ −Fxz) implies (∀x)−Fxx
3. (∀x)(∀y)(∀z)(Fxy.Fyz ⊃ Fxz) and (∀x)−Fxx imply (∀x)(∀y)(Fxy ⊃ −Fyx)
4. (∀x)(∀y)(Fxy ⊃ −Fxx.−Fyy)” is equivalent to (∀x)−Fxx
5. p∨(∀x)Fx and (∀x)(p ∨ Fx) are equivalent

1. Show by deduction that (∀x)(∀y)(Fxy ⊃ Fxx.Fyy) is equivalent to (∀x)[(∃z)Fxz ∨ (∃z)Fzx ⊃ Fxx].
2. Consider the schemata:

   (1) (∀x)(p ≡ Fx) (2) p ≡ (∀x)Fx (3) p ≡ (∃x)Fx  (4) (∃x)(p ≡  Fx)

   Show that:

   1. (1) implies (2) and (1) implies (3).
   2. (2) implies (4) and that (3) implies (4).
3. Schematize the following argument, and show by deduction that the premises imply the conclusion:

   There is a painting which is admired by every critic who admires some painting.
   Every critic admires some painting.
   THEREFORE, some painting is admired by every critic.

1. Find schemata containing the identity sign and no other predicate letters which are ⊤ iff the universe contains:
   1. At least 4 members
   2. Exactly 3 members
   3. No more than 4 and no fewer than 2 members
2. Schematize:
   1. There are exactly three people in the room.
   2. There is only one witness, and she is not talking.
3. Show, by deduction, using the laws of identity, that:
   1. (∃x)(Fx.Rxy) . (∀x)(Fx ⊃ −Rxz) implies y≠ z
   2. (∃x)(∀y)(x=y ⊃ Fxy) is equivalent to (∃x)Fxx

1. Schematize and show by deduction that the following arguments are valid:
   1. All musicians are talented
      Dave is a musician
      Dave is a child
      THEREFORE, some child is talented
   2. Tony likes everyone who talks to him
      No one Alex talks to likes anyone
      THEREFORE, Alex does not talk to Tony
2. Schematize the following argument with the description operator. Then show by deduction, using an additional descriptive premise, that the argument is valid.

   The logic teacher likes some student.
   Curmudgeons are likes no one.
   THEREFORE, curmudgeons don’t teach logic.

3. Schematize the following statements using descriptions. Then eliminate the descriptions by Russell’s method. In (ii), carry out the elimination in two ways, taking the description to have broader or narrower scope.
   1. The skeptic whom no stoic challenges refutes everyone.
   2. The unchallenged skeptic does not refute Chrysippus.
4. Show by deduction with a premise to the effect that “the unchallenged skeptic” is a proper description that the two schemata obtained in (ii) in the last problem are equivalent.
5. Consider the following rule of Limited Universal Generalization, acronym LUG, that is the same as our normal UG, except the variable of quantification must be the same as the variable free in the schema from which one generalizes.  For example, from Fx by LUG one can deduce only (∀x)Fx, not (∀y)Fy.  Show that all applications of UG are eliminable in favor of LUG and the other rules of deduction minus the EI rules.
6. Suppose we add to the full deduction system of Goldfarb a New Rule, acronym NR: if a schema R ⊃ S occurs on a line (m), and u occurs free neither in S nor in any premise of line (m), then on a subsequent line we may put (∃u)R ⊃ S, with the same premise numbers as line (m) and citation (m). Is the expanded deduction system still sound? Why or why not?

Additional Exercises for Test 2:

1. Put each of the following into prenex form, and show by deduction that it is equivalent to its prenex form:
   1. (∃x)Fx ∨(∃y)(∀z)Gyz
   2. (∀x)[(∃y)Rxy ⊃ (∀z)Rzx]
2. Give the first 10 steps of the application of the Rigid Plan to: (∀x)(∃y)(Fxy ⊃ Fyx).