The University of Western Ontario
The University of Western Ontario London Canada Department of Philosophy Philosophy 212 (001) UW – Introduction to Logic Mid-Year Test 3 hours Closed Book
L. Falkenstein Wednesday, December 14, 2011 7:00pm, EC 2155 Page one of one page
Answer All Questions 1. Explicate in terms of open and/or closed truth-trees: 6% a. Truth-functional consistency A set, Γ, of sentences is t-f consistent if and only if the tree for Γ has a completed open branch. (Marks off if you did not specify both “if” and “only if.” Marks off as well if you only said that the tree is open. Unfinished trees are also open.) b. Truth-functional falsity A sentence, P, is t-f false if and only if either the tree for P is closed or (equivalently) every possible truth value assignment can be recovered from the completed tree for ~P. 2. Under what conditions is a tree branch open? A tree branch is open if and only if it is not closed, which is to say, if and only if an atomic sentence, P, and its negation ~P do not occur on the branch. This could happen because the branch has not been finished yet or it could be the case after all non-literal sentences on the branch have been checked off, in which case the branch is not only open but completed.2% 3. Give the first three lines of the tree used to test “A ⊃ B” and “~B ⊃ ~A” for t-f equivalence. 3% Line 1: ~[(A ⊃ B) ≡ (~B ⊃ ~A)] Line 2 left: A ⊃ B Line 2 right: ~(A ⊃ B) Line 3 left ~(~B ⊃ ~A) Line 3 right ~B ⊃ ~A (Points off if the punctuation is not absolutely accurate, esp wrt relative position of “~” and punctuation marks. Points off as well if the sentences are left-right reversed. 4. Use the truth-tree method to answer a and b below. In each case state your result. Where appropriate, recover a truth-value assignment that supports your answer. 6% a. Is the following argument truth-functionally valid? [C ⊃ (A & U)] ≡ (C v A) ~(A & ~B) (B & A) ⊃ (~C v ∼U) CvB
Since the tree for the set of the premises plus the negation of the conclusion is closed, the argument is t-f valid. (Points off for failing to state this.) b. Are the following sentences truth-functionally equivalent? A ⊃ (B ⊃ C) (A ⊃ B) ⊃ C
Since the tree for the negation of the biconditional of the two sentences has at least one completed open branch, the sentences are not t-f equivalent. The truth value assignments recoverable from the leftmost open branch that prove this are: A=F, B=T, C=F and A=F, B=F, C=F. (Points off for failing to include this information. Note that you need not complete the tree if you can find a completed open branch, but that you must do everything on other branches that is required by a rule application to the branch you are working on. E.g., decomposing the sentence at line 2 requires decomposing the sentence on every branch that opens beneath it, not just the ones on the left that you are working on. No sentences are decomposed on the rightmost branch because no rule application requires it. Note also that you need only recover interpretations from one open branch, though you should recover all the interpretations from this branch that come off of it.)
5. Suppose that the completed truth-tree for a sentence P has at least one open branch and that the completed truth-tree for a sentence Q has at least one open branch. Does it follow that the tree for P & Q must have at least one open branch? No. Explain, using an example if appropriate. Consider the tree for “A & ~A.” This tree closes after one step, applying &D. But the (one line long) trees for both “A” and “~A” each have a completed open branch. 4% 6. Define 6% a. Derivability (of a sentence from a set of assumptions) in SD. A sentence, P, is derivable in SD from a set of assumptions, Γ, if and only if there is a derivation in SD that contains sentences from Γ as its initial assumptions and that derives P under the scope just of those assumptions (i.e., not just under the scope of auxiliary assumptions). b. Equivalence (of two sentences) in SD. Two sentences, P and Q are equivalent in SD if and only if Q is derivable from {P} in SD and P is derivable from {Q} in SD. 7. Construct derivations that establish the following: a. ‘A ⊃ ~B’ and ‘B ⊃ ~A’ are equivalent in SD (do not use rules of SD+) 1. A ⊃ ~B 2. B 3. A 4. B 5. ~B 6. ~A 7. B ⊃ ~A
2, R 1, 3, ⊃E 3-4&5, ~I 2-6, ⊃I
B ⊃ ~A A B A ~A ~B A ⊃ ~B
2, R 1, 3, ⊃E 3-4&5, ~I 2-6, ⊃I
1 2 3 4 5 6 7
6%
b. The following argument is valid in SD or SD+ A ⊃ (B & ~C) C v ~B C v ~A 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
A ⊃ (B & ~C) C v ~B C C v ~A
3, vI
~B A B & ~C B ~B ~A C v ~A C v ~A
1, 6, ⊃E 7, &E 5, R 6-8&9, ~I 10, vI 2, 3-4, 5-11, vI
One of these answers is sufficient.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
A ⊃ (B & ~C) C v ~B ~B v C B⊃C A B & ~C B A⊃B A⊃C ~A v C C v ~A
2, Assoc 3, Impl. 1, 5, ⊃E 6, &E 5-7, ⊃I 8, 4, HS 9, Impl. 10, Assoc.
8. Show that the following argument is valid in SD or SD+:
3%
BvM B⊃C (M ⊃ G) & (G ⊃ Z) (E v R) ≡ ~(C v Z) E ⊃ ∼K 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.
BvM B⊃C (M ⊃ G) & (G ⊃ Z) (E v R) ≡ ~(C v Z) E K EvR ~(C v Z) ~C & ~Z ~C ~B M M⊃ G G G⊃Z Z ~Z ~K E ⊃ ~K
5, vI 4, 7, ≡E 8, DeM 9, &E 2, 10, MT 1, 11, DS 3, &E 13, 12, ⊃E 3, ,&E 15, 14, ⊃E 9, &E 6-16&17, ~E 5-18 ⊃I
9. Explain why any argument of SL whose conclusion is a theorem in SD is valid in SD. 4% Suppose the conclusion, P, of an argument in SL is a theorem of SD. Then by definition of “theorem of SD” P is derivable in SD from the empty set. Since P is derivable from the empty set, it will also be derivable from any other set, including the set of assumptions that consists of the premises of the argument of SL that has P as its conclusion, since the same sequence of steps used to derive P from the empty set can be used to derive it under the scope of any set of assumptions whatsoever. But by definition, an argument is valid in SD if its conclusion is derivable from the set of its premises in SD. So if the conclusion, P, of an argument in SL is a theorem of SD, that argument will be valid in SD.
L. Falkenstein Wednesday, December 14, 2011 7:00pm, EC 2155 Page one of one page
Answer All Questions 1. Explicate in terms of open and/or closed truth-trees: 6% a. Truth-functional consistency A set, Γ, of sentences is t-f consistent if and only if the tree for Γ has a completed open branch. (Marks off if you did not specify both “if” and “only if.” Marks off as well if you only said that the tree is open. Unfinished trees are also open.) b. Truth-functional falsity A sentence, P, is t-f false if and only if either the tree for P is closed or (equivalently) every possible truth value assignment can be recovered from the completed tree for ~P. 2. Under what conditions is a tree branch open? A tree branch is open if and only if it is not closed, which is to say, if and only if an atomic sentence, P, and its negation ~P do not occur on the branch. This could happen because the branch has not been finished yet or it could be the case after all non-literal sentences on the branch have been checked off, in which case the branch is not only open but completed.2% 3. Give the first three lines of the tree used to test “A ⊃ B” and “~B ⊃ ~A” for t-f equivalence. 3% Line 1: ~[(A ⊃ B) ≡ (~B ⊃ ~A)] Line 2 left: A ⊃ B Line 2 right: ~(A ⊃ B) Line 3 left ~(~B ⊃ ~A) Line 3 right ~B ⊃ ~A (Points off if the punctuation is not absolutely accurate, esp wrt relative position of “~” and punctuation marks. Points off as well if the sentences are left-right reversed. 4. Use the truth-tree method to answer a and b below. In each case state your result. Where appropriate, recover a truth-value assignment that supports your answer. 6% a. Is the following argument truth-functionally valid? [C ⊃ (A & U)] ≡ (C v A) ~(A & ~B) (B & A) ⊃ (~C v ∼U) CvB
Since the tree for the set of the premises plus the negation of the conclusion is closed, the argument is t-f valid. (Points off for failing to state this.) b. Are the following sentences truth-functionally equivalent? A ⊃ (B ⊃ C) (A ⊃ B) ⊃ C
Since the tree for the negation of the biconditional of the two sentences has at least one completed open branch, the sentences are not t-f equivalent. The truth value assignments recoverable from the leftmost open branch that prove this are: A=F, B=T, C=F and A=F, B=F, C=F. (Points off for failing to include this information. Note that you need not complete the tree if you can find a completed open branch, but that you must do everything on other branches that is required by a rule application to the branch you are working on. E.g., decomposing the sentence at line 2 requires decomposing the sentence on every branch that opens beneath it, not just the ones on the left that you are working on. No sentences are decomposed on the rightmost branch because no rule application requires it. Note also that you need only recover interpretations from one open branch, though you should recover all the interpretations from this branch that come off of it.)
5. Suppose that the completed truth-tree for a sentence P has at least one open branch and that the completed truth-tree for a sentence Q has at least one open branch. Does it follow that the tree for P & Q must have at least one open branch? No. Explain, using an example if appropriate. Consider the tree for “A & ~A.” This tree closes after one step, applying &D. But the (one line long) trees for both “A” and “~A” each have a completed open branch. 4% 6. Define 6% a. Derivability (of a sentence from a set of assumptions) in SD. A sentence, P, is derivable in SD from a set of assumptions, Γ, if and only if there is a derivation in SD that contains sentences from Γ as its initial assumptions and that derives P under the scope just of those assumptions (i.e., not just under the scope of auxiliary assumptions). b. Equivalence (of two sentences) in SD. Two sentences, P and Q are equivalent in SD if and only if Q is derivable from {P} in SD and P is derivable from {Q} in SD. 7. Construct derivations that establish the following: a. ‘A ⊃ ~B’ and ‘B ⊃ ~A’ are equivalent in SD (do not use rules of SD+) 1. A ⊃ ~B 2. B 3. A 4. B 5. ~B 6. ~A 7. B ⊃ ~A
2, R 1, 3, ⊃E 3-4&5, ~I 2-6, ⊃I
B ⊃ ~A A B A ~A ~B A ⊃ ~B
2, R 1, 3, ⊃E 3-4&5, ~I 2-6, ⊃I
1 2 3 4 5 6 7
6%
b. The following argument is valid in SD or SD+ A ⊃ (B & ~C) C v ~B C v ~A 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
A ⊃ (B & ~C) C v ~B C C v ~A
3, vI
~B A B & ~C B ~B ~A C v ~A C v ~A
1, 6, ⊃E 7, &E 5, R 6-8&9, ~I 10, vI 2, 3-4, 5-11, vI
One of these answers is sufficient.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
A ⊃ (B & ~C) C v ~B ~B v C B⊃C A B & ~C B A⊃B A⊃C ~A v C C v ~A
2, Assoc 3, Impl. 1, 5, ⊃E 6, &E 5-7, ⊃I 8, 4, HS 9, Impl. 10, Assoc.
8. Show that the following argument is valid in SD or SD+:
3%
BvM B⊃C (M ⊃ G) & (G ⊃ Z) (E v R) ≡ ~(C v Z) E ⊃ ∼K 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.
BvM B⊃C (M ⊃ G) & (G ⊃ Z) (E v R) ≡ ~(C v Z) E K EvR ~(C v Z) ~C & ~Z ~C ~B M M⊃ G G G⊃Z Z ~Z ~K E ⊃ ~K
5, vI 4, 7, ≡E 8, DeM 9, &E 2, 10, MT 1, 11, DS 3, &E 13, 12, ⊃E 3, ,&E 15, 14, ⊃E 9, &E 6-16&17, ~E 5-18 ⊃I
9. Explain why any argument of SL whose conclusion is a theorem in SD is valid in SD. 4% Suppose the conclusion, P, of an argument in SL is a theorem of SD. Then by definition of “theorem of SD” P is derivable in SD from the empty set. Since P is derivable from the empty set, it will also be derivable from any other set, including the set of assumptions that consists of the premises of the argument of SL that has P as its conclusion, since the same sequence of steps used to derive P from the empty set can be used to derive it under the scope of any set of assumptions whatsoever. But by definition, an argument is valid in SD if its conclusion is derivable from the set of its premises in SD. So if the conclusion, P, of an argument in SL is a theorem of SD, that argument will be valid in SD.
