Section 1:
Chapter 1 Part C
1. consistent
2. inconsistent
3. consistent
4. consistent
Chapter 1 Part D 1, 2, 3, 6, 8, and 10 are possible.
Chapter 2 Part A
1. ¬\(M\)
2. \(M\)∨¬\(M\)
3. \(G\)∨\(C\)
4. ¬\(C\) &¬\(G\)
5. \(C\) → (¬\(G\)&¬\(M\))
6. \(M\)∨(\(C\)∨\(G\))
Chapter 2 Part C
1. \(E\)
1
&\(E\)
2
2. \(F\)
1
→ \(S\)
1
3. \(F\)
1
∨\(E\)
1
4. \(E\)
2
&¬\(S\)
2
5. ¬\(E\)
1
&¬\(E\)
2
6. \(E\)
1
&\(E\)
2
&¬(\(S\)
1
∨\(S\)
2
)
7. \(S\)
2
→ \(F\)
2
8. (¬\(E\)
1
→ ¬\(E\)
2
)&(\(E\)
1
→ \(E\)
2
)
9. \(S\)
1
↔¬\(S\)
2
10. (\(E\)
2
&\(F\)
2
) → \(S\)
2
11. ¬(\(E\)
2
&\(F\)
2
)
12. (\(F\)
1
&\(F\)
2
) ↔ (¬\(E\)
1
&¬\(E\)
2
)
Chapter 2 Part D
A:
Alice is a spy.
B:
Bob is a spy.
C:
The code has been broken.
G:
The German embassy will be in an uproar.
1. \(A\)&\(B\)
2. (\(A\)∨\(B\)) → \(C\)
3. ¬(\(A\)∨\(B\)) →¬\(C\)
4. \(G\)∨\(C\)
5. (\(C\)∨¬\(C\))&\(G\)
6. (\(A\)∨\(B\))&¬(\(A\)&\(B\))
Chapter 2 Part G
1. (a) no (b) no
2. (a) no (b) yes
3. (a) yes (b) yes
4. (a) no (b) no
5. (a) yes (b) yes
6. (a) no (b) no
7. (a) no (b) yes
8. (a) no (b) yes
9. (a) no (b) no
Chapter 3 Part A
1. tautology
2. contradiction
3. contingent
4. tautology
5. tautology
6. contingent
7. tautology
8. contradiction
9. tautology
10. contradiction
11. tautology
12. contingent
13. contradiction
14. contingent
15. tautology
16. tautology
17. contingent
18. contingent
Chapter 3 Part B 2, 3, 5, 6, 8, and 9 are logically equivalent.
Chapter 3 Part C 1, 3, 6, 7, and 8 are consistent.
Chapter 3 Part D 3, 5, 8, and 10 are valid.
Chapter 3 Part E
1. \(\mathcal{A}\) and \(\mathcal{B}\) have the same truth value on every line of a complete truth table, so \(\mathcal{A}\) ↔ \(\mathcal{B}\) is true on every line. It is a tautology.
2. The sentence is false on some line of a complete truth table. On that line, \(\mathcal{A}\) and \(\mathcal{B}\) are true and \(\mathcal{C}\) is false. So the argument is invalid.
3. Since there is no line of a complete truth table on which all three sentences are true, the conjunction is false on every line. So it is a contradiction.
4. Since \(\mathcal{A}\) is false on every line of a complete truth table, there is no line on which \(\mathcal{A}\) and \(\mathcal{B}\) are true and \(\mathcal{C}\) is false. So the argument is valid.
5. Since \(\mathcal{C}\) is true on every line of a complete truth table, there is no line on which \(\mathcal{A}\) and \(\mathcal{B}\) are true and \(\mathcal{C}\) is false. So the argument is valid.
6. Not much. (\(\mathcal{A}\) ∨ \(\mathcal{B}\)) is a tautology if \(\mathcal{A}\) and \(\mathcal{B}\) are tautologies; it is a contradiction if they are contradictions; it is contingent if they are contingent.
7. \(\mathcal{A}\) and \(\mathcal{B}\) have different truth values on at least one line of a complete truth table, and (\(\mathcal{A}\)∨\(\mathcal{B}\)) will be true on that line. On other lines, it might be true or false. So (\(\mathcal{A}\)∨\(\mathcal{B}\)) is either a tautology or it is contingent; it is not a contradiction.
Chapter 3 Part F
1. ¬\(A\) → \(B\)
2. ¬(\(A\) →¬\(B\))
3. ¬[(\(A\) → \(B\)) →¬(\(B\) → \(A\))]
Chapter 4 Part A
1. \(Za\)&\(Zb\)&\(Zc\)
2. \(Rb\)&¬\(Ab\)
3. \(Lcb\) → \(Mb\)
4. (\(Ab\)&\(Ac\)) → (\(Lab\)&\(Lac\))
5. ∃\(x\)(\(Rx\)&\(Zx\))
6. ∀\(x\)(\(Ax\) → \(Rx\))
7. ∀\(x\)[\(Zx\) → (\(Mx\)∨\(Ax\))]
8. ∃\(x\)(\(Rx\)&¬\(Ax\))
9. ∃\(x\)(\(Rx\)&\(Lcx\))
10. ∀\(x\)[(\(Mx\)&\(Zx\)) → \(Lbx\)]
11. ∀\(x\)[(\(Mx\)&\(Lax\)) → \(Lxa\)]
12. ∃\(xRx\) → \(Ra\)
13. ∀\(x\)(\(Ax\) → \(Rx\))
14. ∀\(x\)[(\(Mx\)&\(Lcx\)) → \(Lax\)]
15. ∃\(x\)(\(Mx\)&\(Lxb\)&¬\(Lbx\))
Chapter 4 Part E
1. ¬∃\(xTx\)
2. ∀\(x\)(\(Mx\) → \(Sx\))
3. ∃\(x\)¬\(Sx\)
4. ∃\(x\)[\(Cx\)&¬∃\(yByx\)]
5. ¬∃\(xBxx\)
6. ¬∃\(x\)(\(Cx\)&¬\(Sx\)&\(Tx\))
7. ∃\(x\)(\(Cx\)&\(Tx\))&∃\(x\)(\(Mx\)&\(Tx\))&¬∃\(x\)(\(Cx\)&\(Mx\)&\(Tx\))
8. ∀\(x\)[\(Cx\) →∀\(y\)(¬\(Cy\) → \(Bxy\))]
9. ∀\(x\)(\(Cx\)&\(Mx\)) →∀\(y\)[(¬\(Cy\) &¬\(My\)) → \(Bxy\)])
Chapter 4 Part G
1. ∀\(x\)(\(Cxp\) → \(Dx\))
2. \(Cjp\)&\(Fj\)
3. ∃\(x\)(\(Cxp\)&\(Fx\))
4. ¬∃\(xSxj\)
5. ∀\(x\)[(\(Cxp\)&\(Fx\)) → \(Dx\)]
6. ¬∃\(x\)(\(Cxp\)&\(Mx\))
7. ∃\(x\)(\(Cjx\)&\(Sxe\)&\(Fj\))
8. \(Spe\)&\(Mp\)
9. ∀\(x\)[(\(Sxp\)&\(Mx\)) →¬∃\(yCyx\)]
10. ∃\(x\)(\(Sxj\) &∃\(yCyx\)&\(Fj\))
11. ∀\(x\)[\(Dx\) →∃\(y\)(\(Sxy\) &\(Fy\) &\(Dy\))]
12. ∀\(x\)[(\(Mx\)&\(Dx\)) →∃\(y\)(\(Cxy\) &\(Dy\))]
Chapter 4 Part J
1. ∀\(x\)(\(Cx\) → \(Bx\))
2. ¬∃\(xWx\)
3. ∃\(x\)∃\(y\)(\(Cx\)&\(Cy\) &\(x\) ≠ \(y\))
4. ∃\(x\)∃\(y\)(\(Jx\)&\(Ox\)&\(Jy\) &\(Oy\) &\(x\) ≠ \(y\))
5. ∀\(x\)∀\(y\)∀\(z\)[(\(Jx\)&\(Ox\)&\(Jy\) &\(Oy\) &\(Jz\) &\(Oz\)) → (\(x\) = \(y\)∨\(x\) = \(z∨\)\(y\) = \(z\))]
6. ∃\(x\)∃\(yJx\)&\(Bx\)&\(Jy\) &\(By\) &\(x\) ≠ \(y\) &∀\(z\)[(\(Jz\) &\(Bz\)) → (\(x\) = \(z\)∨\(y\) = \(z\))])
7. ∃\(x\)
1
∃\(x\)
2
∃\(x\)
3
∃\(x\)
4
[\(Dx\)
1
&\(Dx\)
2
&\(Dx\)
3
&\(Dx\)
4
&\(x\)
1
≠ \(x\)
2
&\(x\)
1
≠ \(x\)
3
&\(x\)
1
≠ \(x\)
4
&\(x\)
2
≠ \(x\)
3
&\(x\)
2
≠ \(x\)
4
&\(x\)
3
≠ \(x\)
4
&¬∃\(y\)(\(Dy\) &\(y\) ≠ \(x\)
1
&\(y\) ≠ \(x\)
2
&\(y\) ≠ \(x\)
3
&\(y\) ≠ \(x\)
4
)]
8. ∃\(xDx\)&\(Cx\)&∀\(y\)[(\(Dy\) & \(Cy\)) → \(x\) = \(y\)]&\(Bx\))
9. ∀\(x\)[(\(Ox\)&\(Jx\)) → \(vWx\)]&∃\(x\)[\(Mx\)&∀\(y\)(\(My\) → \(x\) = \(y\))&\(Wx\)]
10. ∃\(xDx\)&\(Cx\)&∀\(y\)[(\(Dy\) & \(Cy\)) → \(x\) = \(y\)]&\(Wx\))→∃\(x\)∀\(y\)(\(Wx\) ↔ \(x\) = \(y\))
11. wide scope: ¬∃\(x\)[\(Mx\)&∀\(y\)(\(My\) → \(x\) = \(y\))&\(Jx\)}
narrow scope: ∃\(x\)[\(Mx\)&∀\(y\)(\(My\) → \(x\) = \(y\))&¬\(Jx\)]
12. wide scope: ¬∃\(x\)∃\(zDx\)&\(Cx\)&\(Mz\) &∀\(y\)[(\(Dy\) & \(Cy\)) → \(x\) = \(y\)]&∀\(y\)[(\(My\) → \(z\) = \(y\))& \(x\) = \(z\)])
narrow scope: ∃\(x\)∃\(zDx\)&\(Cx\)&\(Mz\) &∀\(y\)[(\(Dy\) & \(Cy\)) → \(x\) = \(y\)]&∀\(y\)[(\(My\) → \(z\) = \(y\))& \(x\) ≠ \(z\)])
Chapter 5 Part A 2, 3, 4, 6, 8, and 9 are true in the model.
Chapter 5 Part B 4, 5, and 7 are true in the model.
Chapter 5 Part D
UD = {10,11,12,13}
extension(\(O\)) = {11,13}
extension(\(S\)) = ∅
extension(\(T\)) = {10,11,12,13}
extension(\(U\)) = {13}
extension(\(N\)) = {<11,10>,<12,11>,<13,12>}
Chapter 5 Part E
1. The sentence is true in this model:
UD = {Stan}
extension(\(D\)) = {Stan}
referent(\(a\)) = Stan
referent(\(b\)) = Stan
And it is false in this model:
UD = {Stan}
extension(\(D\)) = ∅
referent(\(a\)) = Stan
referent(\(b\)) = Stan 2.
The sentence is true in this model:
UD = {Stan}
extension(\(T\)) = {<Stan, Stan>}
referent(\(h\)) = Stan
And it is false in this model:
UD = {Stan}
extension(\(T\)) = ∅
referent(\(h\)) = Stan
3. The sentence is true in this model:
UD = {Stan, Ollie}
extension(\(P\)) = {Stan}
referent(\(m\)) = Stan
And it is false in this model:
UD = {Stan}
extension(\(P\)) = ∅
referent(\(m\)) = Stan
Chapter 5 Part F There are many possible correct answers. Here are some:
1. Making the first sentence true and the second false:
UD = {alpha}
extension(\(J\)) = {alpha}
extension(\(K\)) = ∅
referent(\(a\)) = alpha
2. Making the first sentence true and the second false:
UD = {alpha, omega}
extension(\(J\)) = {alpha}
referent(\(m\)) = omega
3. Making the first sentence false and the second true:
UD = {alpha, omega}
extension(\(R\)) = {<alpha,alpha>}
4. Making the first sentence false and the second true:
UD = {alpha, omega}
extension(\(P\)) = {alpha}
extension(\(Q\)) = ∅
referent(\(c\)) = alpha
5. Making the first sentence true and the second false:
UD = {iota}
extension(\(P\)) = ∅
extension(\(Q\)) = ∅
6. Making the first sentence false and the second true:
UD = {iota}
extension(\(P\)) = ∅
extension(\(Q\)) = {iota}
7. Making the first sentence true and the second false:
UD = {iota}
extension(\(P\)) = ∅
extension(\(Q\)) = {iota}
8. Making the first sentence true and the second false:
UD = {alpha, omega}
extension(\(R\)) = {<alpha, omega>, <omega, alpha>}
9. Making the first sentence false and the second true:
UD = {alpha, omega}
extension(\(R\)) = {<alpha, alpha>, <alpha, omega>}
Chapter 5 Part I
1. There are many possible answers. Here is one:
UD = {Harry, Sally}
extension(\(R\)) = {<Sally, Harry>}
referent(\(a\)) = Harry
2. There are no predicates or constants, so we only need to give a UD. Any UD with 2 members will do.
3. We need to show that it is impossible to construct a model in which these are both true. Suppose∃\(x\) \(x\) ≠ \(a\) is true in a model. There is something in the universe of discourse that is not the referent of \(a\). So there are at least two things in the universe of discourse: referent(\(a\)) and this other thing. Call this other thing \(β\)— we know \(a\) ≠ \(β\). But if \(a\) ≠ \(β\), then ∀\(x\)∀\(y\) \(x\) = \(y\) is false. So the first sentence must be false if the second sentence is true. As such, there is no model in which they are both true. Therefore, they are inconsistent.
Chapter 5 Part J
2. No, it would not make any difference. The satisfaction of a formula with one or more free variables depends on what the variable assignment does for those variables. Because a sentence has no free variables, however, its satisfaction does not depend on the variable assignment. So a sentence that is satisfied by some variable assignment is satisfied by every other variable assignment as well.
Chapter 6 Part A