question archive Prove the following in SD+ #1
Prove the following in SD+ #1
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Prove the following in SD+
#1.
1. R ⊃ (~A & T)
2. B ∨ ~S
3. R ∨ ~S / A ⊃ B
#2.
(A ⊃ A) ⊃ (~A & ~A) / A ∨ ~A
#3.
1. A ∨ B
2. C
3. (A & C) ⊃ D / D ∨ B
#4.
[(A ∨ B) & (D & F)] ∨ [(A ∨ B) & C] / C ∨ F
#5 Show that the following pair of sentence is equivalent in SD+
(A & B) ∨ [(C & D) ∨ A]
([(C ∨ A) & (C ∨ B)] & [(D ∨ A) & (D ∨ B)]) ∨ A
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#1
1. R ⊃ (~A & T)
2. B ∨ ~S
3. R ∨ ~S / A ⊃ B
4. ~R ∨ (~A & T) 1. Material Implication
5. ~R ? (A & T) 4. De Morgan's Theorem
6. (A & T) 5. Simplification
7. A 6. Simplification
8. B ? ~S 2. De Morgan's Theorem
9. B 8. Simplification
10. A ⊃ B 7, 9 Material Implication
#2
1. (A ⊃ A) ⊃ (~A & ~A) / A ∨ ~A
2. ~ (A ⊃ A) ∨ (A & A) 1. Material Implication
3. ~ (A ⊃ A) ? (A & A) 2. De Morgan's Theorem
4. (A & A) 3. Simplification
5. A 4. Simplification
6. ~ (A ⊃ A) 3. Simplification
7. ~ A ∨ ~A 6. Material Implication
8. ~ A ? ~A 7. De Morgan's Theorem
9. ~ A 8. Simplification
10. A ∨ ~A 5, 6 Disjunction
#3
1. A ∨ B
2. C
3. (A & C) ⊃ D / D ∨ B
4. ~ (A & C) ∨ D 3. Material Implication
5. (~ A & ~ C) ? D 4. De Morgan's Theorem
6. D 5. Simplification
7. A ? B 1. De Morgan's Theorem
8. B 7. Simplification
9. D ∨ B 6, 8 Disjunction
#4
1. [(A ∨ B) & (D & F)] ∨ [(A ∨ B) & C] /C ∨ F
2. [(A ∨ B) & (D & F)] ? [(A ∨ B) & C] 1. De Morgan's Theorem
3. [(A ∨ B) & C] 2. Simplification
4. C 3. Simplification
5. [(A ∨ B) & (D & F)] 2. Simplification
6. (D & F) 5. Simplification
7. F 6. Simplification
8. C ∨ F 4, 7 Disjunction
#5
(A & B) ∨ [(C & D) ∨ A]
1. (A & B) ∨ [(C & D) ∨ A]
2. (A & B) ? [(C & D) ∨ A] 1. De Morgan's Theorem
3. [(C & D) ∨ A] 2. Simplification
4. [(C & D) ? A] 3. De Morgan's Theorem
5. (C & D) 4. Simplification
6. A 4. Simplification
([(C ∨ A) & (C ∨ B)] & [(D ∨ A) & (D ∨ B)]) ∨ A
1. ([(C ∨ A) & (C ∨ B)] & [(D ∨ A) & (D ∨ B)]) ∨ A
2. [(C ∨ A) & (C ∨ B)] 1. Simplification
3. (C ∨ A) 2. Simplification
4. C ? A 3. De Morgan's Theorem
5. A 4. Simplification
-The above pair of sentences are logically equivalent because they entail each other.
Step-by-step explanation
-The following arguments are proved using SD to show that they are valid.
#1
1. R ⊃ (~A & T)
2. B ∨ ~S
3. R ∨ ~S / A ⊃ B
4. ~R ∨ (~A & T) 1. Material Implication
5. ~R ? (A & T) 4. De Morgan's Theorem
6. (A & T) 5. Simplification
7. A 6. Simplification
8. B ? ~S 2. De Morgan's Theorem
9. B 8. Simplification
10. A ⊃ B 7, 9 Material Implication
#2
1. (A ⊃ A) ⊃ (~A & ~A) / A ∨ ~A
2. ~ (A ⊃ A) ∨ (A & A) 1. Material Implication
3. ~ (A ⊃ A) ? (A & A) 2. De Morgan's Theorem
4. (A & A) 3. Simplification
5. A 4. Simplification
6. ~ (A ⊃ A) 3. Simplification
7. ~ A ∨ ~A 6. Material Implication
8. ~ A ? ~A 7. De Morgan's Theorem
9. ~ A 8. Simplification
10. A ∨ ~A 5, 6 Disjunction
#3
1. A ∨ B
2. C
3. (A & C) ⊃ D / D ∨ B
4. ~ (A & C) ∨ D 3. Material Implication
5. (~ A & ~ C) ? D 4. De Morgan's Theorem
6. D 5. Simplification
7. A ? B 1. De Morgan's Theorem
8. B 7. Simplification
9. D ∨ B 6, 8 Disjunction
#4
1. [(A ∨ B) & (D & F)] ∨ [(A ∨ B) & C] /C ∨ F
2. [(A ∨ B) & (D & F)] ? [(A ∨ B) & C] 1. De Morgan's Theorem
3. [(A ∨ B) & C] 2. Simplification
4. C 3. Simplification
5. [(A ∨ B) & (D & F)] 2. Simplification
6. (D & F) 5. Simplification
7. F 6. Simplification
8. C ∨ F 4, 7 Disjunction
#5
(A & B) ∨ [(C & D) ∨ A]
1. (A & B) ∨ [(C & D) ∨ A]
2. (A & B) ? [(C & D) ∨ A] 1. De Morgan's Theorem
3. [(C & D) ∨ A] 2. Simplification
4. [(C & D) ? A] 3. De Morgan's Theorem
5. (C & D) 4. Simplification
6. A 4. Simplification
([(C ∨ A) & (C ∨ B)] & [(D ∨ A) & (D ∨ B)]) ∨ A
1. ([(C ∨ A) & (C ∨ B)] & [(D ∨ A) & (D ∨ B)]) ∨ A
2. [(C ∨ A) & (C ∨ B)] 1. Simplification
3. (C ∨ A) 2. Simplification
4. C ? A 3. De Morgan's Theorem
5. A 4. Simplification
-Noticeably, the above pair of sentences are logically equivalent because they entail each other in the sense that the truth value of (A & B) ∨ [(C & D) ∨ A] /A is TRUE and the truth value of ([(C ∨ A) & (C ∨ B)] & [(D ∨ A) & (D ∨ B)]) ∨ A/A is also TRUE.
