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10–21
Determine which of the lettered claims below is equivalent to the following: The gun can
be sold even though it has no trigger lock. (This can be easy to do if you symbolize the
claims first and have some familiarity either with truth tables or with the Group II rules for
derivations—the truth-functional equivalences.)
A. Only if the gun has a trigger lock can it be sold.
B. The gun has no trigger lock, but it can be sold anyway.
C. If the gun cannot be sold, then it has no trigger lock.
D. If the gun has no trigger lock, then it can be sold.
Equivalent to B
Use the short truth-table method to determine whether the following is valid or invalid:
P → Q
Q → R
R → S
P/∴ S v T
Valid
Use the short truth-table method to determine whether the following is valid or invalid:
A → B
C → D
B v D/∴ A v C
Invalid: A = F
C = F
Make either B or D = T
Use the short truth-table method to determine whether the following is valid or invalid:
A → B
~D → ~C
~D
~C → ~B/∴ ~A
Valid.
Use the short truth-table method to determine whether the following is valid or invalid:
A → B
~C v D
E → F
G → H
A v C
E v G/∴ (B v D) v (F v H)
Valid.
Use the short truth-table method to determine whether the following is valid or invalid:
A → B
C → D
~(~A & ~C)/∴ ~B → (D & ~B)
Valid.
Use the short truth-table method to determine whether the following is valid or invalid:
~E v ~D
~(A → C)
~(~B v ~E)
~(A & B) v (C → D)/∴ ~A
Invalid: A = T
B = T
C = F
D = F
E = T
Use the short truth-table method to determine whether the following is valid or invalid:
P → Q
~Q v R
~(R & ~S)
~P/∴ ~S
Use the short truth-table method to determine whether the following is valid or invalid:
P → (Q & R)
R → (T v S)
~T v (U & ~W)
P/∴ ~W
Invalid: P = T
Q = T
R = T
S = T
T = F
U = T or F
W = T
Use the short truth-table method to determine whether the following is valid or invalid:
P → Q
~(Q & ~R)
~R v S
~(S & ~T)
T/∴ P
Invalid: P = F OR P = F OR P = F
Q = F Q = T Q = F
R = F R = T R = T or F
S = F S = T S = T
T = T T = T T = T
Use the short truth-table method to determine whether the following is valid or invalid:
M → (N → O)
(~N v O) → ~P
P/∴ ~M
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. P → (Q & R) (Premise)
2. (Q & R) → S (Premise)/∴ P → S
3. P
→ S 1,2, CA
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. P → Q (Premise)
2. P v R (Premise)
3. R → (S & T) (Premise)/∴ Q v (S & T)
4. Q v (S & T) 1,3,2, CD
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. (P & Q) → (R v S) (Premise)
2. (R v S) → T (Premise)
3. P & Q (Premise)/∴ T
4. R v S 1,3, MP
5. T 2,4, MP
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. (R v Q) & S (Premise)
2. (R v Q) → ~P (Premise)
3. P v T (Premise)/∴ T
4. R v Q 1, SIM
5.
~P 2,4, MP
6. T 3,5, DA
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. P → (S v T) (Premise)
2. S → Q (Premise)
3. P (Premise)
4. T → R (Premise)/∴ Q v R
5. S v T 1,3, MP
6.
Q v R 2,4,5, CD
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. ~P & Q (Premise)
2. R → P (Premise)/ ∴ Q & ~R
3.
~P 1, SIM
4.
~R 2,3, MT
5. Q 1, SIM
6. Q &
~R 4,5, CONJ
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. P → M (Premise)
2. ~M v ~Q (Premise)
3. S → Q (Premise)
4. (~P v ~S) → R (Premise)/∴ R
5. ~P v ~S 1,3,2, DD
6. R 4,5, MP
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. Q → (P → R) (Premise)
2. ~R & Q (Premise)/∴ ~P
3. Q 2, SIM
4. P
→
R 1,3, MP
5. ~R 2, SIM
6. ~P 4,5, MT
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
1. (P v Q) → (X & Z) (Premise)
2. (X → W) & (Z → Y) (Premise)
3. P (Premise)/∴ W & Y
4. P v Q 3, ADD
5
.
X & Z 1,4, MP
6
.
X 5, SIM
7
.
X
→
W 2, SIM
8
.
W 7,6, MP
9
.
Z 5, SIM
10. Z
→
Y 2, SIM
11
.
Y 10,9, MP
12
.
W & Y 8,11, CONJ
Using only rules from Group I, provide a deductive proof demonstrating the validity of the
following:
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P & Q (Premise)
2. R v ~(P & Q) (Premise)/∴ R
3.
~~(P & Q) 1, DN
4
.
R 2,3, DA
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. ~(P v Q) (Premise)/∴ ~Q
2.
~P &
~Q 1, DEM
3. ~Q 2, SIM
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. (P v Q) & (P v R) (Premise)
2. ~P (Premise)/∴ Q & R
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P v Q (Premise)
2. Q → R (Premise)/∴ ~P → R
3.
~P
→
Q 1, IMPL
4. ~P
→
R 2,3, CA
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P → Q (Premise)
2. R → Q (Premise)
3. P v R (Premise)/∴ Q
4. Q v Q 1,2,3 CD
5. Q 4, TAUT
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P v ~Q (Premise)
2. R → Q (Premise)/∴ R → P
3.
~Q v P 1, COM
4. Q
→
P 3, IMPL
5. R
→
P 2,4, CA
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P v (Q & R) (Premise)
2. Q → ~R (Premise)/∴ P
3.
~Q v
~R 2, IMPL
4.
~(Q & R) 3, DEM
5. P 1,4, DA
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. (M v P) → Q (Premise)
2. Q → (~L v R) (Premise)
3. (L → R) → S (Premise)/∴ (M v P) → S
4. (M v P)
→
(~L v R) 1,2, CA
5. (M v P)
→
(L
→ R) 4, IMPL
6. (M v P)
→
S 3,5, CA
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P → ~(M & R) (Premise)
2. ~P → Q (Premise)/∴ ~Q → (~M v ~R)
This item shows the advantage of the deductive method for proving validity over even the
short truth-table method. Only a modest familiarity with CA and IMPL should enable you
to do this problem in your head.
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P v (Q & P) (Premise)
2. P → R (Premise)/∴ R
3.
(P v Q) & (P v P) 1, DIST
4.
P v P 3, SIM
5.
P 4, TAUT
6.
R 2,5, MP
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P → (Q → R) (Premise)/∴ Q → (P → R)
2. (P & Q)
→ R 1, EXPORT
3.
(Q & P)
→ R 2, COM
4.
Q
→ (P
→ R) 3, EXPORT
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. (P → Q) & (R → S) (Premise)
2. Q → ~S (Premise)
3. ~T → (P & R) (Premise)/∴ T
4. P
→ Q 1, SIM
5.
R
→ S 1, SIM
6.
~Q v
~S 2, IMPL
7.
~P v
~R 4,5,6, DD
8.
~(P & R) 7, DEM
9. T 3,8, MT
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P → (Q & R) (Premise)
2. ~S → ~(Q v R) (Premise)/∴ P → S
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P → (Q & R) (Premise)
2. R → (S v T) (Premise)/∴ ~(~P v S) → T
3.
~P v (Q & R) 1, IMPL
4.
~P v (R & Q) 3, COM
5.
(~P v R) & (~P v Q) 4, DIST
6.
~P v R 5, SIM
7. P
→ R 6, IMPL
8. P
→ (S v T) 2,7, CA
9. P
→ (~S
→ T) 8, IMPL
10. (P & ~S)
→ T 9, EXPORT
11. (~~P & ~S)
→ T 10, DN
12. ~(~P v S)
→ T 11, DEM
Using rules from both Group I and Group II, construct a deduction to prove that the
following is valid.
1. P → (Q & R) (Premise)
2. ~(S & Q) (Premise)
3. (R → S) & Q (Premise)/∴ ~P
4.
~S v
~Q 2, DEM
5.
~Q v
~S 4, COM
6.
Q
→ ~S 5, IMPL
7. Q 3, SIM
8.
~S 6,7, MP
9. R
→ S 3, SIM
10. ~R 9,8, MT
11. ~R v ~Q 10, ADD
12. ~(R & Q) 11, DEM
13. ~(Q & R) 12, COM
14. ~P 1,14, MT
Determine whether the following symbolized argument is valid or invalid. If invalid, provide
a counterexample; if valid, construct a deduction.
P v (R & ~S)
Q → ~P/∴ Q → S
Invalid.
P = F
Q = T
R = T
S = F
Determine whether the following symbolized argument is valid or invalid. If invalid, provide
a counterexample; if valid, construct a deduction.
P → (Q v R)
~S v ~T
(Q v R) → (S & T)/∴ ~P