- #1
TheAlli
Hi! I was kind of struggling with a couple of the problems on my symbolic logic homework, and any help/hints/etc. would be very much appreciated!
The same symbolizations are used as in Klenk's "Understanding Symbolic Logic" book:
• = dot, meaning "and", as in p • q
v = wedge, meaning "either/or", as in p v q
≡ = triple bar, meaning "if and only if", as in p ≡ q
~ = tilde, meaning "not", as in ~p
⊃ = horseshoe, meaning "if ___, then ___", as in p ⊃ q
The inference rules (rules that must apply to an entire line in any proof created) we've learned so far are Modus Ponens, Modus Tollens, Hypothetical Syllogism, Simplification, Conjunction, Dilemma, Disjunctive Syllogism, and Addition. The replacement rules (rules that may apply to only part of a line in any proof created) we've learned are Double Negation, Duplication, Commutation, Association, Contrapostion, DeMorgan's, Biconditional Exchange, Conditional Exchange, Distribution, and Exportation. A summary of the rules is given here: http://www.yuksel.org/e/philosophy/logic/19rules.htm
The first problem I was having issues with presents this premise and conclusion (the Pr. indicating that I justify making that line by it being a premise of the given problem, with the conclusion following the slash):
1. (A ⊃ C) Pr. / ((A • B) ⊃ C)
Since I was told that working backward is a good tactic, I started with the conclusion and noted that, by the exportation rule, it is equivalent to ((A ⊃ (B ⊃ C)). Because B must be added on by the rule of addition, I must be able to prove that either A or C, or maybe ~A/~C is true. I tried using contraposition to get ((A ⊃ (~C ⊃ ~B)), then changing to ((A • ~C) ⊃ ~B), but I'm not sure how to use Modus Tollens and get (A • ~C) by itself since that would require getting ~~B, which can't happen as far as I know without getting one of the other variables anyway. So I'm stuck in a bit of a vicious cycle.
Another problem that's giving me issues is:
1. (A ⊃ ~A) Pr. / ~A
I thought using conditional exchange would be good here, since you could start from the premise and get (~A v ~~A) or get the same by working backward from the conclusion using Conditional Exchange, and this would be equivalent to (~A v A) due to Double Negation. From there I tried DeMorgan's and got ~(A • ~A), but it's sort of "trapped" within the tilde even though I know replacement rules can still apply to any individual part (for example, I could still change it to ~(~~A • ~A) by Double Negation). Basically, I need to get A/~A by itself and somehow that should work out a proof that contradicts itself.
One other thing, if I may; I completed this problem but am not sure if I'm misusing Conditional Exchange, and if I knew it could maybe help on the previous problem:
1. A Pr. / ~(A ⊃ ~B)
2. B Pr.
3. ~~A DN 1
4. ~~B DN 2
5. ~~A • ~~B Conj. 1, 2
6. ~(~A v ~B) DeM 5
7. B v ~A Add. 2
8. ~B DS 3, 7
7. ~(A ⊃ ~B) CE 6
Since Conditional Exchange allows going from (~p v q) to (p ⊃ q), I know usually ~(~A v ~B) would not translate into the conditional because the substitution doesn't fit the form without assuming that ~B stands for q while A substitutes for p. Is this fine as long as ~B has been proven?
Again, thanks so much in advance and sorry for making my first post a homework-related one. Symbolic logic is definitely not my strong suit!
EDIT: Sorry about the spacing issue on the proofs, for some reason I can't put more than one space between the line and the justification.
The same symbolizations are used as in Klenk's "Understanding Symbolic Logic" book:
• = dot, meaning "and", as in p • q
v = wedge, meaning "either/or", as in p v q
≡ = triple bar, meaning "if and only if", as in p ≡ q
~ = tilde, meaning "not", as in ~p
⊃ = horseshoe, meaning "if ___, then ___", as in p ⊃ q
The inference rules (rules that must apply to an entire line in any proof created) we've learned so far are Modus Ponens, Modus Tollens, Hypothetical Syllogism, Simplification, Conjunction, Dilemma, Disjunctive Syllogism, and Addition. The replacement rules (rules that may apply to only part of a line in any proof created) we've learned are Double Negation, Duplication, Commutation, Association, Contrapostion, DeMorgan's, Biconditional Exchange, Conditional Exchange, Distribution, and Exportation. A summary of the rules is given here: http://www.yuksel.org/e/philosophy/logic/19rules.htm
The first problem I was having issues with presents this premise and conclusion (the Pr. indicating that I justify making that line by it being a premise of the given problem, with the conclusion following the slash):
1. (A ⊃ C) Pr. / ((A • B) ⊃ C)
Since I was told that working backward is a good tactic, I started with the conclusion and noted that, by the exportation rule, it is equivalent to ((A ⊃ (B ⊃ C)). Because B must be added on by the rule of addition, I must be able to prove that either A or C, or maybe ~A/~C is true. I tried using contraposition to get ((A ⊃ (~C ⊃ ~B)), then changing to ((A • ~C) ⊃ ~B), but I'm not sure how to use Modus Tollens and get (A • ~C) by itself since that would require getting ~~B, which can't happen as far as I know without getting one of the other variables anyway. So I'm stuck in a bit of a vicious cycle.
Another problem that's giving me issues is:
1. (A ⊃ ~A) Pr. / ~A
I thought using conditional exchange would be good here, since you could start from the premise and get (~A v ~~A) or get the same by working backward from the conclusion using Conditional Exchange, and this would be equivalent to (~A v A) due to Double Negation. From there I tried DeMorgan's and got ~(A • ~A), but it's sort of "trapped" within the tilde even though I know replacement rules can still apply to any individual part (for example, I could still change it to ~(~~A • ~A) by Double Negation). Basically, I need to get A/~A by itself and somehow that should work out a proof that contradicts itself.
One other thing, if I may; I completed this problem but am not sure if I'm misusing Conditional Exchange, and if I knew it could maybe help on the previous problem:
1. A Pr. / ~(A ⊃ ~B)
2. B Pr.
3. ~~A DN 1
4. ~~B DN 2
5. ~~A • ~~B Conj. 1, 2
6. ~(~A v ~B) DeM 5
7. B v ~A Add. 2
8. ~B DS 3, 7
7. ~(A ⊃ ~B) CE 6
Since Conditional Exchange allows going from (~p v q) to (p ⊃ q), I know usually ~(~A v ~B) would not translate into the conditional because the substitution doesn't fit the form without assuming that ~B stands for q while A substitutes for p. Is this fine as long as ~B has been proven?
Again, thanks so much in advance and sorry for making my first post a homework-related one. Symbolic logic is definitely not my strong suit!
EDIT: Sorry about the spacing issue on the proofs, for some reason I can't put more than one space between the line and the justification.
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