Twootie Conversions
The truth tree system in Twootie matches closely with Meaning and Argument, but not exactly. The notes below should make switching between the two easier.
Corresponding Logics
Meaning and Argument  Twootie 
Propositional Logic (PL)  Sentential Problems 
Property Predicate Logic (PPL)  No real equivalent* 
Relational Predicate Logic (RPL)  Predicate Logic Problems 
Relational Predicate Logic with Identity (RPL=)  Identity/Function Problems 
*You could try the first few Predicate Logic problems, if you don't mind having variables in there!
Symbol Changes
Meaning and Argument  Twootie  
Negation  ~  ¬ 
Conditional  ⊃  > 
Biconditional*  ( ⊃ ) & ( ⊃ )  ≡ 
Negation of Identity  ≠  ¬ = 
Variables**  x, y, z, x1,...x2, …  w, x, y ,z 
Predicates*** 
A^{1}, B^{1}, C^{1}, … A^{2}, … A^{3}, … 
A, B, C, … 
*The biconditional (If and Only If) can be written in terms of two conjoined conditionals, thus Meaning and Argument does not assign it a new symbol (see section 7.10). Thus, "P ≡Q " is equivalent to "(P ⊃Q) & (Q ⊃P )".
**Variables in Twootie cannot take subscripts.
***Twootie predicates do not take place numbers.
Terminology
Meaning and Argument  Twootie 
Singular terms  Constants 
Unbound variables  Free variables 
Truth Tree Rules
Meaning and Argument  Twootie 
N/A  All rules end in "d" or "D" 
Branches done by hand  Automatic Branching: Just enter the first branch's contents and the rule. 
Quantifier Exchange (QE) 
Either ¬∀∃ 
Universal Quantifier (UQ)  ∀ 
Existential Quantifier (EQ)  ∃ 
Identity Out (IO)  = 
Identity In (II)  No rule  branch can be closed if it contains, e.g., "Ø a=a" 
Biconditionals: see below. 
Biconditionals
As M&A treats biconditionals as conjoined conditionals, there are no special rules for dealing with them (& and ⊃ rules are enough). Twootie's bicondional (≡) is a new symbol requiring its own rules:
P ≡ Q Ø(P ≡ Q) /\ / \ P ¬P P ¬P Q ¬Q ¬Q Q
Habits
Twootie doesn't let you take the short cuts that you probably do on paper. Meaning and Argument supports these short cuts, so Twootie will not let you replicate some of the book's solutions exactly. The differences aren't substantial though.
 You must decompose the entire sentence once you've started it. Thus this, e.g., is unacceptable:
1) ~P Premise
2) P & Q Negation of conclusion
3) P 2, &
X
Twootie requires a step 4 before the tree can close:
4) Q 2, &
X
 Double negations must be removed explicitly with the double negation rule (~~ in M&A, ¬ ¬ in Twootie). Meaning and Argument introduces this rule, but it's pretty easy to skip in practice. It's required by Twootie, though.
 Rules can't be applied multiple times in one step. This shows up most frequently with Quantifier Exchanges.Meaning and Argument treats the following inference as acceptable:
 ~(∀x)(∃y)(Q2xy)
 (∃x)(∀y)~(Q2xy) 1, QE twice
Twootie won't let you do this, though. The rule must be applied to the first quantifier, then it must be instantiated before the rule can apply to the second quantifier. For example:
 ¬ (∀x)(∃y)(Qxy)
 (∃x)¬ (∃y)(Qxy) 1, ¬ ∀ d (i.e. QE)
 ¬ (∃y)(Qay) 2, ∃d (i.e. EQ)
 (∀y)¬ (Qay) 3, ¬ ∃d (i.e. QE)
M&A lets you perform all the quantifier exchanges before instantiating; with Twootie, you have to start instantiating before completing the exchanges.
 The Identity Out rule in Twootie (=) works a little differently than in M&A. You can only put the first singular term in place of the second, but not vice versa. So a=b let's us substitute a for b, but not b for a.
Completing the trees
If a branch closes, put an "X" under it. If it is open but complete, hit F10 then use the branch menu to declare it open.
Solutions  RPL=
Chapters 16 & 17
Chapter 16
A.
#1.
 (∀x) (B^{1}x ⊃ R^{2}hx)
 h = a
 (∀x) (B^{1}x ⊃ R^{2}ax)
#2.
 (∀x) (D^{1}x ⊃ P^{2}cx)
 c = d
 (∀x) (D^{1}x ⊃ P^{2}dx)
B.
1.  (∀x) (F^{2}xa ⊃ (B^{1}x & C^{1}x))  P  
2.  F^{2}ba  P  
X  3.  ~(∃x) (B^{1}x & C^{1}x & x=b)  NC  
4.  (∀x) ~(B^{1}x & C^{1}x & x=b)  3, QE  
X  5.  F^{2}ba ⊃ (B^{1}b & C^{1}b)  1, UQ  
X  6. 
/ \ ~F^{2} ba (B^{1} b & C^{1} b) X  
5, ⊃  
7.  B^{1}b  6, &  
8.  C^{1}b  
X  9.  ~(B^{1}b & C^{1}b & b=b)  4, UQ  
10. 
/  \ ~B^{1} b ~C^{1} b ~b=b X X  
9, ~& twice  
11. 
b=b X 
I I 
All branches closed: valid argument
C.
 (∃x) (P^{1}x & (∃y) (P^{1}y & x ≠y & H^{1}x & H^{1}y))
 (∀x) (T^{1}x & D^{1}x ⊃ (∀y) (T^{1}y & D^{1}y ⊃ x = y))
D.
 (∃x) (P^{1}x & L^{2}xs & (∀y) ((P^{1}y & L^{2}ys) ⊃ x=y))
 (∃x) (P^{1}x & (∀y) (C^{1}y ⊃ E^{2}xy) & (∀z) ((P^{1}x & (∀x_{1}) (C^{1}x_{1}⊃ E^{2}zx_{1})) ⊃ x_{1} = z))
E.
#1.
 (∃x) (M^{1}x & L^{2}xl & (∀y) ((M^{1}y & L^{2}yl) ⊃ x = y) & L^{1}x)
 (∃x) (M^{1}x & L^{1}x & (∀y) ((M^{1}y & L^{1}y) ⊃ x=y)
#2.
 (∀x) ((P^{1}x & E^{1}x) ⊃ (∀y) ((P^{1}y & E^{1}y) ⊃ (∀z)((P1z & E1z) ⊃ (x = y v x=z v y=z))))
 E^{1}n & E^{1}l
 ~E^{1}m
#3.
 (∃x) (H^{1}x & H^{2}ax & (∃y) (H^{1}y & H^{2}ay & x≠y))
 (H^{1}c & H^{2}ac) & (H^{1}k & H^{2}ak)
 c = k ⊃ (∃x) (H^{1}x & H^{2}ax & x ≠c & x≠k)
Chapter 17
A.
#1.
 (∃x) (T^{2}xc & I^{2}xr & A^{2}xt)
 (∃x) (∃y) (P^{1}y & T^{2}xy & A^{2}xt)
#2.
 (∃x) (K^{3}xaj &A^{2}xw)
 (∃x) (∃y) (∃z) (P^{1}y & P^{1}z & K^{3}xyz & A^{2}xw)
B.
#1.
 (∃x) (S^{2}xm & B^{1}x) & (∃y) (A^{2}ym & R^{1}y)
 (∃x) (S^{2}xm) & (∃y) (A^{2}ym)
#2.
 (∃x) (P^{2}xr & H^{1}x) & (∃y) (W^{2}yr & Q^{1}y)
 ~(∃x) (P^{2}xr & Q^{1}x)
#3.
 (∃x) (P^{1}x & (∃y) (E^{2}yx & (∀z) (D^{1}z ⊃ O^{2}yz)))
 (∀x) (E^{2}xa ⊃ (O^{2}xs v O^{2}xw))
 (∃x) (P^{1}x & (∃y) (E^{2}yx) & x≠a)
C.
#1.
 (∀x) (∃y) ((P^{1}y & T^{2}xy & R^{1}x) ⊃ (∃z) (P^{1}z & z≠y & (∃x_{1}) (M^{3}x_{1}zy)))
 (∃x) (T^{2}xl & R^{1}x) ⊃ (∃y) (P^{1}y & (∃z) (M^{3}zyl))
#2.
 (∃x) (∃y) (∃z) (A^{3}xyz & B^{1}x) ⊃ (∃x_{1}) (∃y_{1}) (P^{1}x_{1} & C^{2}y_{1}x_{1})
 (∃x) (∃y) (∃z) (A^{3}xyz & L^{1}x) ⊃ (∃x_{1}) (∃y_{1}) (P^{1}x_{1} & L^{2}y_{1}x_{1} & H^{1}y_{1})
 (∃x) (A^{3}xja) ⊃ ((∃y) (∃z) (P^{1}y & C^{2}zy) v (∃x_{1}) (∃y_{1}) (P^{1}x_{1} & L^{2}y_{1}x_{1}))
Solutions  PPL and PL
Chapters 7, 911
Chapter 7
A.
 B ⊃ L
 R ⊃ J
 L ⊃ F
 O ⊃ W
 B ⊃ F
 B ⊃ J
B.

J ⊃ R
 R ⊃ A
 ~H ⊃ L
 L ⊃ T
 ~P ⊃ S
 ~~C ⊃ G
C.
 (R ⊃ W) & (W ⊃ R)
 S ∴ ~T
 D ⊃( C ⊃ E)
 (O ⊃ C) & (C ⊃ O)
 (O ⊃ C) & (C ⊃ O)
D.
P  Q  (P & Q) ⊃ (P v Q) 
T  T 
T T T 
F  T 
F T T 
T  F 
F T T 
F  F 
F T F 
P  Q  R  (P v Q) v (~P v R) 
T  T  T 
T T T 
F  T  T 
T T T 
T  F  T 
T T T 
F  F  T 
F T T 
T  T  F 
T T F 
F  T  F 
T T T 
T  F  F 
T T F 
F  F  F 
F T T 
Chapter 9
A.
 K^{1}s
 ∃T^{1}
 ∀G^{1}
 F^{1j}
 W^{1l}
 ∀S^{1}
 ∃R^{1}
B.
 Something is a sheep
 Something is a yellow dog.
 Something is either tall, or a sheep.
Chapter 10
A.
#1.
X  1.  ∃(S^{1} & J^{1})  P 
2.  ∀(J^{1} ⊃ L^{1})  P  
X  3.  ~∀(S^{1} ⊃ L^{1)}  NC 
X  4.  ∃~(S^{1} ⊃ L^{1)}  3, QE 
X  5.  ~(S^{1}a ⊃L^{1}a)  4, EQ 
X  6.  S^{1}b & J^{1}b  1, EQ 
X  7.  J^{1}a ⊃ L^{1}a  2, UQ 
X  8.  J^{1}b ⊃ L^{1}b  2, UQ 
9.  S^{1}a  5,~⊃  
10.  ~L^{1}a  
11.  S^{1}b  6, &  
12. 
J^{1} b 

13. 
/ \ ~J^{1} a L^{1} a  X 
7, ⊃  
14. 
/ \~J^{1} b L^{1} b X O 
8, ⊃ 
Open branches: invalid argument
#2.
X  1.  M^{1}a & N^{1}a  P 
X  2.  ∃(M^{1} & J^{1})  P 
X  3.  ~∃(J^{1} & N^{1})  NC 
4.  ∀~(J^{1} & N^{1})  3, QE  
X  5.  M^{1}b & J^{1}b  2, EQ 
X  6.  ~(J^{1}a & N^{1}a)  4, UQ 
X  7.  ~(J^{1}b & N^{1}b)  4, UQ 
8.  M^{1}a  1, &  
9.  N^{1}a  
10.  M^{1}b  5, &  
11. 
J^{1} b 

12. 
/ \~J^{1} a ~N^{1} a  X 
6, ~&  
13. 
/ \~J^{1} b ~N^{1} b X O 
7, ~& 
Open branches: invalid argument
#3.
X  1.  K1a & B1b  P 
X  2.  ∃(K1 & D1)  P 
3.  ∀(D1 ⊃ B1)  P  
X  4.  ~(B1a v K1b)  NC 
X  5.  K1c & D1c  2, EQ 
X  6.  D1a ⊃ B1a  3, UQ 
X  7.  D1b ⊃ B1b  3, UQ 
X  8.  D1c ⊃ B1c  3, UQ 
9.  K1a  1, &  
10.  B1b  
11.  K1c  5, &  
12.  D1c  
13.  ~B1a  4, ~v  
14. 
~K1b 

15. 
/ \~D^{1} a B^{1} a  X 
6, ⊃  
16. 
/ \~D^{1} c B^{1} c X  
8, ⊃  
17. 
/ \ ~D^{1} b B^{1} b O O 
7, ⊃ 
Open branches: invalid argument
Chapter 11
 ∀((P^{1} & S^{1}) ⊃ L^{1})
 ∃(P^{1} & M^{1}) ⊃ M^{1}l
 ∀(F^{1} ⊃ C^{1})
 ∀ ((P^{1} & T^{1}) ⊃ D ^{1})
 ∃(P^{1} & L^{1}) ⊃ ∀(P^{1} ⊃ L^{1})
Solutions  RPL
Chapters 1215
Chapter 12
A.
 (∀x) (P^{1}x ⊃ L^{2}xr)
 (∃x) (L^{2}jx) v (∃y (F^{2}jy)
 (∃x) (P^{1}x & S^{2}jx)
 (∃x) (M^{1}x & S^{2}dx)
 (∀x) (S^{1}x ⊃ T^{3}axt)
B.
#1.
 (∀x) (B^{1}x ⊃ H^{2}rx)
 B^{1}f
 ∴ H^{2}rf
#2.
 (∃x) (C^{1}x & B^{3}nxr)
 ∴ (∃x) (C^{1}x & H^{2}rx)
#3.
 K^{3}ceb
 ∴(∃x) (P^{1}x & K^{3}cxb) & (∃y) (P^{1}y & K^{3}yeb)
Chapter 13
A.
 (∃x) (P^{1}x & (∀y) (P^{1}y ⊃ L^{2}xy))
 (∃x) (P^{1}x & (∀y) (P^{1}y ⊃ L^{2}yx))
 (∃x) (M^{1}x & (∀y) (B^{1}y ⊃ (∀z) (L^{1}z ⊃ P^{3}yzx)))
 (∃x) (G^{1}x & ~(∃y) (B^{1}y & (∀z) (P^{1}z ⊃ P^{3}xyz)))
 (∀x) (D^{1}x ⊃ (∃y) (S^{1}y & (∃z) (P^{1}z & W^{3}yxz)))
 (∃x) (B^{1}x & (∃y) (S^{1}y & (∀z) (T^{1}z ⊃ R^{3}yxz)))
 (∃x) (C^{1}x & ∀y) (P^{1}y ⊃∀z) (D^{1}z ⊃ S^{3}yxz)))
B.
#1.
 (∀x) (L^{1}x ⊃ (∃y) (F^{1}y & (∃z) (T^{1}z & P^{3}xyz)))
 (∃x) (L^{1}x)
 ∴ (∃x) (F^{1}x & (∃y) (T^{1}y & (∃z) (L^{1}z & P^{3}zxy)))
#2.
 (∃x) (P^{1}x & (∀y)) (T^{1}y ⊃ (∃z) (F^{1}z & E^{3}xzy)))
 ∴ (∀x)) (T^{1}x ⊃ (∃y) (F^{1}y & (∃z) (P^{1}z & E^{3}zyx)))
#3.
 (∃x) (T^{1}x & (∀y) (C^{1}y ⊃ (∃z) (G^{1}z & ~C^{3}yxz))) (This sentence is ambiguous; if you think the sentence can be true if one or two cats climb the tree  as long as not ALL cats do so  then you might be inclined to symbolize it as follows: (∃x) (T^{1}x & ~(∀y) (C^{1}y ⊃ (∃z) (G^{1}z & C^{3}yxz))) )
 ∴ (∀x) (C^{1}x ⊃ (∃y) (G^{1}y & (∃z) (T^{1}z & C^{3}xzy)))
#4.
 (∀x) (L^{1}x ⊃ (∃y) (T^{1}y & (∀z) (W^{1}z ⊃ L^{3}yxz)))
 (∃x) (L^{1}x)
 ∴(∃x) (L^{1}x & (∀y) (W^{1}y ⊃ (∃z) (T^{1}z & C^{3}zxy)))
Chapter 14
A.
#1.
1. 
(∀x) (B^{1}x ⊃H^{2}rx) 
P  
2.  B^{1}f  P  
3.  ~H^{2}rf  NC  
X  4.  B^{1}f ⊃ H^{2}rf  1, UQ 
5. 
/ \ ~B^{1} f H^{2} rf X X 
4, ⊃ 
All branches close: valid argument.
#2.
X  1. 
(∃x) (C^{1}x & B^{3}nxr) 
P 
X  2.  ~(∃x) (C^{1}x & H^{2}rx)  NC 
3.  (∀x) ~ (C^{1}x & H^{2}rx)  2, QE  
X  4.  C^{1}a & B^{3}nar  1, EQ 
X  5.  ~ (C^{1}a & H^{2}ra)  3, UQ 
6.  C^{1}a  4, &  
7.  B^{3}nar  
8. 
/ \ ~C^{1} a ~H^{2} ra X O 
5, ~& 
Open branch: invalid argument
Chapter 15
A.
 ~(∃x) (P^{1}x & (∀y) (T^{1}y ⊃ H^{2}xy))
 ~(∃x) (P^{1}x & ~(∃y) (T^{1}y & H^{2}xy))
 (∃x) (C^{1}x & (∀y) (G^{1}y ⊃ ~E^{2}xy))
 (∀x) (C^{1}x ⊃ ~(∃y) (G^{1}y & E^{2}xy))
B.
#1.
 (∀x) (H^{1}x ⊃ (∃y) (P^{1}y & ~(∃z) (N^{1}z & C^{3}xyz)))
 (∀x) (H^{1}x ⊃ A^{1}x)
 ∴ ~( (∀x) (A^{1}x ⊃ (∃y) (P^{1}y & (∃z) (P^{1}z & C^{3}xyz))))
#2.
 ~(∃x) (D^{1}x & B^{1}x)
 (∀x) (D^{1}x v A^{1}x)
 (∃x) (P^{1}x & ~A^{1}x)
 ∴~(~(∃x) (P^{1}x & ~B^{1}x))
C.
 (∀x) ( (∃y) (F^{1}y & H^{2}xy) ⊃ C^{1}x)
 (∀x) ((∃y) (W^{1}y & B^{2}xy) ⊃ F^{1}x)
 (∀x) (P^{1}x ⊃ ((∀y) (B^{2}xy ⊃ A^{1}y))
 ~((∀x) ((∀y) (S^{2}xy ⊃ B^{1}y ) ⊃ S^{1}x)
D.
#1.
 (∀x) ((∃y) (T^{1}y & C^{2}xy) ⊃ C^{1}x)
 (∀x) ((∃y) (S^{1}y & A^{2}xy) ⊃ (∃z) (T^{1}z & C^{2}xz))
 ∴ (∀x) ((∃y) (S^{1}y & A^{2}xy) ⊃ C^{1}x)
#2.
 (∀x) ((∀y) (L^{2}xy ⊃ L^{1}y) ⊃ T^{1}x)
 (∃x) (P^{1}x & (∃y) (B^{1}y & L^{2}xy))
 ∴ ~(∀x) (P^{1}x ⊃ T^{1}x)
E.
#1
 (∀x) (P^{1}x ⊃ ((∀y) (C^{1}y ⊃ (O^{2}xy ⊃ B^{2}xy))))
 (∀x) ((C^{1}x & (∃y) (P^{1}y & B^{2}yx)) ⊃ S^{1}x)
 (∀x) ((C^{1}x & (∃y) (P^{1}y & B^{2}yx)) ⊃S^{1}x)
#2.
 (∀x) (B^{1}x ⊃ ((∀y) (A^{1}y ⊃ (M^{2}xy ⊃ R^{2}xy))))
 B^{1}s & A^{1}h
 M^{2}sh ⊃ R^{2}sh
#3.
 (∀x) (P^{1}x ⊃ ((∀y) (F^{1}y ⊃ (S^{2}xy ⊃ K^{2}xy))))
 ~(∀x) (F^{1}x ⊃ ((∀y) (P^{1}y ⊃ (K^{2}yx ⊃ S^{2}yx))))
 (∃x) (P^{1}x & (∃y) (F^{1}y & S^{2}xy & ~K^{2}xy))
F.
#1#4: see M&A solutions
#5.
 (∀x) (C^{1}x ⊃ ((∀y) (M^{1}y ⊃ (C^{2}xy ⊃ E^{2}xy))))
 (∀x) ((∃y) (M^{1}y & E^{2}xy) ⊃ ~H^{1}x)
 ~(∃x) (C^{1}x & (∃y) (M^{1}y & C^{2}xy) & H^{1}x)
#6.
 (∀x) ((P^{1}x & F^{1}x) ⊃ (∀y) (P^{1}y ⊃ L^{2}yx))
 (∀x) (P^{1}x ⊃ (∃y) (P^{1}y & ~L^{2}yx)
 ~(∃x) (P^{1}x & F^{1}x)
#7.
 ~(∃x) (P^{1}x & (∃y) (D^{1}y & P^{2}xy) & U^{1}x)
 ~(∀x) (P^{1}x ⊃ (∃y) (D^{1}y & P^{2}xy))
 ~ (∀x) (P^{1}x ⊃ (∃y) ((D^{1}y &P^{2}xy) ⊃ ~U^{1}x))
#8.
 (S^{1}j ⊃ (∀x) (B^{1}x ⊃ P^{1}x)) & ((∀y) (B^{1}y ⊃ P^{1}x) ⊃ S^{1}j)
 (∃x) (T^{1}x & M^{2}jx) ⊃ (∀x) (B^{1}x ⊃ ~P^{1}x)
 ~(∃x) (T^{1}x & M^{2}jx) ⊃ A^{1}j
 ~A^{1}j ⊃ ~S^{1}j
#9.
 ~(∃x) (P^{1}x & (∀y) (T^{1}y ⊃ (~S^{2}xy & W^{2}xy))
 ~(∃x) (P^{1}x & (∀y) (T^{1}y ⊃ ~W^{2}xy) & (∀z) (C^{1}z ⊃ W^{2}xz))
 (∀x) ((P^{1}x & (∀y) (C^{1}y ⊃ W^{2}xy)) ⊃ (∀z) (T^{1}z ⊃ S^{2}xz))
New Exercises
Chapters:  1  2  3  4  5 6  7  8  9  10  11  12  13  14  15  16  17 
 Solutions for PL and PPL (Chapters 7, 911)
 Solutions for RPL (Chapters 1215)
 Solutions for RPL= (Chapters 16, 17)
The best way to learn logic is through practice; below are additional exercises for the more difficult chapters. Follow the links above for solutions.
Chapter 7
When to try: After sect 7.5 Necessary and Sufficient Conditionals, p.88
A. Statements
 Symbolize (1)(5) in PL using the dictionary provided.
1) If the book gets published, then Lucy will throw a party. (B: The book gets published; Lucy throws a party)
2) If Rachel gets an ‘A’, Joan will be pleased. (R: Rachel gets an A; J: Joan is pleased)
3) Mary will feed you if she likes you. (F: Mary feeds you; L: Mary likes you)
4) Wearing clothes is necessary for going outside. (W: you wear clothes; O: you go outside)
5) Bribing the judge is sufficient for going free. (B: you bribe the judge; F: You go free)
6) John’s cooking dinner is necessary for Bill’s being happy. (J: John cooks dinner; B: Bill is happy)
When to try: After sect 7.7 Unless, p.92
B. Statements
 Symbolize (1)(6) in PL using the dictionary provided.
1) Joe will wear is hat only if it rains. (J: Joe wears his hat; R: it rains)
2) Susan will be rich only if her aunt dies. (R: Susan will be rich; A: Susan’s aunt dies)
3) I will live to be 100 unless I get hit by a sugarcrazed New York City taxi driver. (L: I will live to be 100; H: I get hit by a sugarcrazed New York City taxi driver)
4) Laura will get an ‘A’ only if she buys the teacher an apple. (L: Laura gets an 'A'; T: Laura buys the teacher an apple)
5) Jerry will be sad unless his paper gets published. (S: Jerry is sad; P: Jerry's paper gets published)
6) Unless I didn’t count correctly, there are three hundred cabbages in this garden. (C: I counted correctly; G: there are three hundred cabbages in this garden)
When to try: After sect 7.10 If and Only If, p.94
C. Statements
 Symbolize (1)(5) in PL using the dictionary provided.
1) Nadia relaxes if and only if all her work is done. (R: Nadia relaxes; W: All Nadia’s work is done)
2) Since I went to the store, I will not run out of toothpaste. (S: I went to the store; T: I will run out of toothpaste)
3) If you eat your dinner, then if there’s cake in the house, you can eat desert. (D: you eat your dinner; C: there’s cake in the house; E: you can eat desert)
4) You can go out if and only if you clean your room. (O: You can go out; C: you clean your room)
5) Cleaning your room is necessary and sufficient for going out.
When to try: Add to Ex. 6, p. 108
D. Truth Tables
 Determine if the following statements are contradictions, tautologies, or contingent using the truth table method.
6. (P & Q) (P v Q)
7. (P v Q) v (~P v R)
When to try: Add to Ex.7, p.109
E. Truth Tables
 Determine whether the set of statements is consistent
3. {(P (Q & R)) & S, ~(T v R) & (S P)}
Chapter 9
When to try: After 9.4 Quantifiers, p. 140
A. Statements
 Symbolize (1)(7) in PPL using the dictionary provided.
1) Susan is kind. (K^{1}: is kind; s: Susan)
2) Something tastes sweet. (T^{1}: tastes sweet)
3) Everything is good. (G^{1}: is good)
4) John is funny. (F^{1}: is funny; j: John)
5) Lucy works hard. (W^{1}: works hard; l: Lucy)
6) Everything smiles a lot. (S^{1}: smiles a lot)
7) Something is rooster. (R^{1}: is rooster)
When to try: add to Ex 1, p. 143
B. Representations
 What English sentences might these represent?
6. ∃S^{1}
7. ∃(D^{1} & Y^{1})
8. ∃(T^{1 }v S^{1})
Chapter 10
When to try: Add to Ex. 2, p. 167
A. Truth Trees
 Construct truth trees for the following.
11) ∃(S^{1} & J^{1}), ∀(J^{1}⊃ L^{1}), ∴∀(S^{1}⊃ L^{1})
12) M^{1}a & N^{1}a, ∃ (M^{1} & J^{1}), ∴∃ (J^{1} & N^{1})
13) K^{1}a & B^{1}b, ∃(K^{1} & D^{1}), ∀(D^{1}⊃ B^{1}), ∴B^{1}a v K^{1}b
Chapter 11
When to try: After 11.4 Pronouns Revisited, p. 178
A. Statements
 Symbolize (1)(5) in PPL using the dictionary provided.
1) Anyone who can sing is lucky. (S^{1}: can sing; L^{1}: is lucky)
2) If anyone will be a millionaire, Laura will be a millionaire. (M^{1}: is a millionaire; l: Laura)
3) If something is frozen, it is cold. (F^{1}: is frozen; C^{1}: is cold)
4) If someone touches the gold, he will die. (P^{1}: is a person; T^{1}: touches the gold; D^{1}: dies)
5) If anyone can laugh, then everyone can laugh. (P^{1}: is a person; L^{1}: can laugh)
Chapter 12
When to try: After Chapter 12, p. 204
A. Statements
 Symbolize (1)(5) in RPL using the dictionary provided.
1) Everyone likes Rebecca. (P^{1}:is a person; L^{2}:likes; r: Rebecca)
2) Either John lost something, or John found something. (L^{2}:lost; F^{2}:found; j: John)
3) Joan sang to someone. (P^{1}:is a person; S^{2}: sang to; j: Joan)
4) There’s a mat that Domino the cat sat on. (S^{2}:sits on; M^{1}: is a mat; d: Domino the cat)
5) Alan teaches every student on Tuesday. (T^{3}:_teaches_on _; S^{1}: is a student; a:Alan; t:Tuesday)
When to try: After Chapter 12, p. 204
B. Arguments
 Symbolize arguments (1)(3) in RPL using the dictionary provided.
1) Rachel has every Beanie Baby. Flip is a Beanie Baby. So, Rachel has Flip. (H^{2}has; B^{1}:is a Beanie Baby; r:Rachel; f:Flip)
2) Nadia bought her roommate a chicken. Hence, Nadia’s roommate has a chicken. (n:Nadia; r:Nadia’s roommate; C^{1}: is a chicken; B^{3}: _bought_for_; H^{2}:has)
3) Carly kissed Erik in Boston. So, Carly kissed someone in Boston and someone kissed Eric in Boston. (K^{3}:_kissed_in_; c: Carly; e:Erik; b:Boston)
Chapter 13
When to try: Change Ex. 5, p. 225. Don't worry about #5, 6, 7, 8, 9, 10, or 11  try these instead!
A. Statements
5) Someone likes everyone.
6) Someone is liked by everyone.
7) Some mailbox is such that every boy put every letter in it. [P^{3}:_put_ into_; B^{1}: is a boy; L^{1}: is a letter; M^{1}: is a mailbox]
8) Some girl didn’t push a boy into every puddle. [P^{3}:_push_ into_; G^{1}: is a girl; P^{1}: is a puddle]
9) Every dog had some student walk it in a park. [W^{3}:_walked_in_; D^{1} is a dog; S^{1}: is a student; P^{1}: is a park]
10) Some book has a student reading it at all times. [R^{3}:_ reads_at _; B^{1}: is a book; T^{1}: is a time]
11) There’s a conundrum that every professor struggled over every day. [S^{3}:_ struggled over_on_; C^{1}: is a conundrum; P^{1}: is a professor; D^{1}: is a day]
When to try: Instead of Ex. 6, p. 225
B. Arguments
1) Every living organism must perform a life function at some time. There are living organisms. Therefore, there’s some life function which at some time has a living organism performing it. [P^{3}:_ performs_at_; L^{1}: is a living organism; F^{1}: is a life function; T^{1}: is a time]
2) Someone is always eating some food. So, at all times, some food is being eaten by someone. [E^{3}:_eats_at_; P^{1}: is a person; F^{1}: is a food; T^{1}: is a time]
3) At least one tree isn’t climbed up by every cat in a garden. So, for every cat, there’s a garden in which there’s a tree that she climbs up. [C^{3}:_ climbs up _in_; T^{1}: is a tree; C^{1}: is a cat; G^{1}: is a garden]
4) Every leaf is lost by some tree every winter. There are leafs. Thus, some leaf is such that every winter some tree loses it. [L^{3}:_loses_in_; L^{1}:is a leaf; W^{1}:is a winter; T^{1}:is a tree]
Chapter 14
When to try: After Chapter 14, p.238
A. Truth Trees
 Determine the validity of the arguments you symbolized in Chapter 12 additional exercises.
1) Rachel has every Beanie Baby. Flip is a Beanie Baby. So, Rachel has Flip. (H^{2}has; B^{1}:is a Beanie Baby; r:Rachel; f:Flip)
2) Nadia bought her roommate a chicken. Hence, Nadia’s roommate has a chicken. (n:Nadia; r:Nadia’s roommate; C^{1}: is a chicken; B^{3}: _bought_for_; H^{2}:has)
3) Carly kissed Erik in Boston. So, Carly kissed someone in Boston and someone kissed Erik in Boston. (K^{3}:_kissed_in_; c: Carly; e:Erik; b:Boston)
Chapter 15
When to try: Add to Ex. 1, p. 240
A. Statements
3) No one is happy always. [H^{2}: is happy at; T^{1}: is a time; P^{1}: is a person]
4) No one is never happy.
5) Some cow doesn’t eat grass. [E^{2}:eats; C^{1}: is a cow; G^{1}: is grass]
6) Every cow eats no grass.
When to try: Add to Ex 2, p. 240
B. Arguments
3) All hydrogens have a proton, but not a neutron. All hydrogens are atoms. So, it’s not true that all atoms have protons and neutrons. [C^{3}:_have_and_; H^{1}: is a hydrogen; P^{1}: is a proton; N^{1}: is a neutron; A^{1}: is an atom]
4) Nothing dead breathes. Everything is either dead or alive. Some people are not alive. So it’s false that no one doesn’t breathe.
When to try: Add to Ex. 3, p. 242
C. Statements
1) Only cats have fur. [H^{2}:_has_; C^{1}: is a cat; F^{1}: is fur]
2) Only fish can breathe under water. (B^{2}: can breathe under; F^{1}: is a fish; W^{1}: is water]
3) Everyone can breathe only air. [B^{2}:can breathe; P^{1}: is a person; A^{1}: is air]
4) Not only students study only books. [S^{2}:studies; S^{1}: is a student; B^{1}: is a book]
When to try: Add to Ex 4, pp.242243
D. Arguments
3) Only cats climb trees. Only tree climbers catch squirrels. So, only cats catch squirrels. [C^{2}:climbs; A^{2}: catches; C^{1}: is a cat; S^{1}: is a squirrel; T^{1}: is a tree]
4) Only trees lose only leaves. Some people lose books. Thus, not every person is a tree. [L^{2}: loses; T^{1}: is a tree; L^{1}: is a leaf: B^{1}: is a book: P^{1}: is a person]
When to try: Add to Ex. 11, p.254
E. Arguments
3) Everyone who owns a cat brushes it. Each cat that is brushed by someone is silky. Therefore, every owned cat is silky. [O^{2}:_owns_; B^{2}:_brushes_; P^{1}: is a person; C^{1}; is a cat; S^{1}: is silky]
4) Any strong base that mixes with a strong acid reacts with it. Sodium hydroxide is a strong base and hydrogen chloride is a strong acid. Thus, if sodium hydroxide mixes with hydrogen chloride, it will react with it. [M^{2}:_mixes with_; R^{2}:_reacts with_; A^{1}: is a strong acid; B^{1}: is a strong base; s: sodium hydroxide; h: hydrogen chloride]
5) Anyone that sits on a flea kills it. But not every flea that is killed by someone is sat on (by him). So, there’s someone who sat on a flea and didn’t kill it. [P^{1}: is a person; S^{2}:sits on; K^{2}:_kills; F^{1}: is a flea]
When to try: After Chapter 15, p. 257
F. Arguments
1) P. 254 Ex 11 #1
2) p. 254 Ex 11 #2
3) p. 225 Ex 6 #1
4) p. 225 Ex 6 #2
5) Whenever a cat catches a mouse, he’ll eat it. Mouse eaters will not go hungry. So, no cat that catches a mouse goes hungry. [C^{2}: catches; C^{1}: is a cat; M^{1}: is a mouse; E^{2}: eats; H^{1}: goes hungry]
6) Only people who are loved by everyone are perfect. Everyone has someone who doesn’t love him. Thus, it follows that nobody is perfect. [L^{2}: loves; P^{1}: is a person; F^{1}: is perfect]
7) No one is unhappy who plays with a dog. Not everyone has a dog he plays with. Thus, it’s not the case that everyone who plays with a dog is not unhappy. [P^{1}: is a person; U^{1}: is unhappy; P^{2}:plays with; D^{1}: is a dog]
8) John is successful if and only if all his books are published. If he made typos, his books will not be published. Being accurate is necessary for not making typos. Therefore, if John is not accurate, he will not be successful. [S^{1}: is successful; P^{1}: is published; B^{1}: is a book of John; j:John; T^{1}: is a typo; M^{2}:made; A^{1}: is accurate]
9) No one who does not study for tests will do well on them. No one who doesn’t do well on tests will do well in courses. So, only people who study for tests will do well in courses. [P^{1}: is a person; T^{1}: is a test; S^{2}:studies for; C^{1}: is a course; W^{2}:does well in]
Chapter 16
When to try: Add to Exercise 1, pp.260261
A. Arguments
3) Hydrogen chloride reacts with every base. Hydrogen chloride is the same as hydrochloric acid. So, hydrochloric acid reacts with every base. (R^{2}: reacts with; B^{1}: is a base; h: hydrogen chloride; a: hydrochloric acid)
4) Alan’s cat purrs everyday. Alan’s cat is Domino. Thus, Domino purrs everyday. (c: Alan’s cat; d: Domino; D^{1}: is a day; P^{2}:purrs on)
When to try: Add to Exercise 3, p. 264
B. Truth Trees
6. (∀x)(F^{2}xa ⊃ (B^{1}x & C^{1}x)); F^{2}ba; ∴(∃x)(B^{1}x & C^{1}x & x = b)
When to try: Before Ex. 5, p. 273
C. Numerical Adjectives
 Solve for (1)(4), using the dictionary provided.
1) At least two people are happy. (P^{1}: is a person; H^{1}: is happy)
2) No more than one turkey will die. (T^{1}: is a turkey; D^{1}: will die)
When to try: After Ex. 6, p. 274.
D. Statements
 Symbolize in RPL^{=}, using the dictionary provided.
1) Susan is loved by exactly one person (L^{2}: loves; s: Susan; P^{1}:is a person)
2) Exactly one person ate every cabbage. (E^{2}: eats; C^{1}: is a cabbage.)
When to try: After Chapter 16, p. 282
E. Arguments
 Symbolize in RPL^{=}, using the dictionary provided.
1) The only man who loves Lucy left. So, exactly one man left. (L^{2}: loves; L^{1}: left; M^{1}: is a man; l: Lucy)
2) No more than two people are eligible. Nadia is eligible, and so is Laura. Therefore, Mike is not eligible. (E^{1}:is eligible; P^{1}: is a person; n: Nadia; l: Laura; m: Mike)
3) Anna has at least two husbands. Craig is her husband and Ken is her husband. Thus, if Craig and Ken are the same person, Anna has another husband. (H^{2}:has; H^{1}: is a husband; c: Craig; k: Ken)
Chapter 17
When to try: After 17.3, p. 293
A. Arguments
 Symbolize in RPL^{=}, using the dictionary provided.
1) Carly talked in her room at 3:00 am. So, someone was talking at 3:00 am. (T^{2}:is a talking by; I^{2}: occurs in; A^{2}: occurs at; c: Carly; t:3:00 am; r: Carly's room)
2) Alex killed John last week. Thus, someone killed someone last week. (K^{3}:_is a killing by_ of _; A^{2}: occurs at; a: Alex; j: John; w: last week)
When to try: After 17.4 Adverbial Modification, p. 295
B. Arguments
 Symbolize arguments (1)(3) into RPL^{=}, using the dictionary provided.
1) Maria sang beautifully and acted brilliantly. So, Maria sang and acted. (S^{2}: is a singing by; A^{2}: is an acting by; B^{1}: is beautiful; R^{1}: is brilliant; m: Maria)
2) Rachel played happily and worked quietly. So, it’s not the case that Rachel played quietly. (P^{2}: is a playing by; W^{2}: is a working by; H^{1}: is happy; Q^{1}:is quiet; r: Rachel)
3) Someone exercises everyday. Alan exercises only on Saturday and Wednesday. Therefore, there is someone who exercises that is not Alan. (P^{1}:is a person; E^{2}: is an exercising by; O^{2}:occurs on; H^{1}:is happy; a: Alan; D^{1}: is a day; O^{2}: occurs on; s: Saturday; w: Wednesday)
When to try: Add to Exercise 2, pp. 297298.
C. Arguments
 Symbolize, using the conventions recommended by the event approach, arguments (1)(2) in RPL^{=}.
1) Whenever someone talks rapidly, someone else misunderstands him. So, if Lisa talks rapidly, someone will misunderstand her. (T^{2}:is a talking by; M^{3}:_is a misunderstanding by_of _; R^{1}:is rapid; P^{1}:is a person; l:Lisa)
2) If there is a big argument, someone will cry. If there is a little argument, someone will laugh heartily. So, if Jerry argues with Alison, then either someone will cry or someone will laugh. (A^{3}:_is an arguing by_with_; B^{1}:is big; L^{1}:is little; H^{1}:is hearty; C^{2}:is a crying by; L^{2}:is a laughing by)