Phi 270 F00 test 3 in pdf format
Analyze the sentences below in as much detail as possible using connectives; that is, you should not identify components that are individual terms (or predicates or functors). Present the result in both symbolic and English notation. Be sure that the unanalyzed components of your answer are complete and independent sentences; also try to respect any grouping in the English.
1. If it rains, you will get wet if you're outside
[answer]
2. Al missed breakfast only if he overslept
[answer]
Use derivations to check whether each of the entailments below holds. You may use detachment and attachment rules. If an entailment fails, present a counterexample that divides an open gap.
3. A → (B → C) ⇒ (A → ¬ C) → (A → ¬ B)
[answer]
4. A → B ⇒ ¬ A ∧ B
[answer]
Analyze the sentence below in as much detail as possible. In this case you should identify components that are individual terms, predicates, or functors. Be sure that the unanalyzed components of your answer are independent (in particular, that none contains a pronoun whose antecedent is in another).
5. Unless Al is the file's owner, the system didn't let him open it
[answer]
[The following question was on a topic not covered in F04] Expand the following sentence in all possible ways on each of the terms appearing in it (i.e., you need not use vacuous abstraction).
6. Tabc
[answer]
Use a derivation to show that the entailment below holds. You may use detachment and attachment rules.
7. A → Ra(fb), Rb(fa) → Ga ⇒ A → (¬ Gb → ¬ a=b)
[answer]
Phi 270 F00 test 3 answers
1. it will rain → you will get wet if you're outside
it will rain → (you will get wet ← you will be outside)
R → (W ← O) [or: R → (O → W)]
if R then if O then W
[O: you will be outside; R: it will rain; W: you will get wet]
2. ¬ Al missed breakfast ← ¬ Al overslept
¬ M ← ¬ O [or: ¬ O → ¬ M)]
if not O then not M
[M: Al missed breakfast; O:Al overslept]
3.
│A → (B → C) 3
├─
││A → ¬ C 4
│├─
│││A (3),(4)
││├─
3 MPP │││B → C 5
4 MPP │││¬ C (5)
5 MTT │││¬ B (6)
│││●
││├─
6 QED │││¬ B 2
│├─
2 CP ││A → ¬ B 1
├─
1 CP │(A → ¬ C) → (A → ¬ B)
4.
│A → B 3,5
├─
│││A (3)
││├─
3 MPP │││B
│││○ A, B ⇏ ⊥
││├─
│││⊥ 2
│├─
2 RAA ││¬ A 1
│
│││¬ B (5)
││├─
5 MTT │││¬ A
│││○ ¬ A, ¬ B ⇏ ⊥
││├─
│││⊥ 4
│├─
4 IP ││B 1
├─
1 Cnj │¬ A ∧ B
A B A → B / ¬ A ∧ B
T T   T F F divides 1st gap
F F   T T F divides 2nd gap
5. ¬ Al is the file's owner → the system didn't let Al open the file
¬ Al is the file's owner → ¬  the system let Al open the file
¬ Al = the file's owner → ¬ [λxyz (x let y open z)] the system Al the file
¬ a = [λx (x's owner)] the file → ¬  Lsaf
¬ a = of → ¬  Lsaf
[L: λxyz (x let y open z); a: Al; f: the file; o: λx (x's owner); s: the system]
6. [This question was on a topic not covered in F04]
[λx Txbc]a
[λx Taxc]b
[λx Tabx]c
7.
│A → Ra(fb)
│Rb(fa) → Ga 4
├─
││A
│├─
││Ra(fb) (5)
││
│││¬ Gb (6)
││├─
││││a=b a=b, fa=fb
│││├─
│││││●
││││├─
5 QED= │││││Rb(fa) 4
││││
│││││Ga (6)
││││├─
│││││●
││││├─
6 Nc= │││││⊥ 4
│││├─
4 RC ││││⊥ 3
││├─
3 RAA │││¬ a=b 2
│├─
2 CP ││¬ Gb → ¬ a=b 1
├─
1 CP │A → (¬ Gb → ¬ a=b)