Exercise 1.4.2
S1. P v ~R, ~R → S, ~P |- S
Our goal is to get the atomic statement S. If we look over the
premises, we find S as the right side of the conditional ~R → S.
If we had the left side of this conditional, ~R, then we could get
S by →E. So, the next step is to get ~R. In fact, ~R occurs in
the premise P v ~R. Since this is a disjunction, we could get ~R by vE if
we had a denial of the other disjunct, P. But we do have this as the
third premise. So, reversing this, we first get ~R by vE, then get
S by →E:
| 1 | (1) | P v ~R | A |
| 2 | (2) | ~R → S | A |
| 3 | (3) | ~P | A |
| 1,3 | (4) | ~R | 1,3 vE |
| 1,2,3 | (5) | S | 2,4 →E |
S2. P v ~R, ~R → S, ~P |- S & ~R
This has the same premises as the previous examle, but now our goal is
a conjunction: S & ~R. We can get a conjunction by &I if we
can get both conjunctions. The previous proof shows how to get S, so
we can use that again here. But in addition, we got ~R as one of the
lines of that previous proof. So, we can get what we want by just
adding one further line to the last proof:
| 1 | (1) | P v ~R | A |
| 2 | (2) | ~R → S | A |
| 3 | (3) | ~P | A |
| 1,3 | (4) | ~R | 1,3 vE |
| 1,2,3 | (5) | S | 2,4 →E |
| 1,2,3 | (6) | S & ~R | 4,5 &I |
S3. P → ~Q, ~Q v R → ~S, P & T |- ~S
Since the conclusion we want, ~S, appears as the consequent of one of
the premises, ~Q v R → ~S, we can get it if we can get the antecedent
of that premise. The antecedent ~Q v R is a disjunction, and we can
get a disjunction by vI if we can get either disjunct. R doesn't occur
anywhere else in the premises, but ~Q is the consequent of P →~Q; so,
if we can get P, we can get what we want. P is one of the conjuncts of
P & T, and we can get it gy &E. So, reversing this reasoning:
| 1 | (1) | P → ~Q | A |
| 2 | (2) | ~Q v R → ~S | A |
| 3 | (3) | P & T | A |
| 3 | (4) | P | 3 &E |
| 1,3 | (5) | ~Q | 1,4 →E |
| 1,3 | (6) | ~Q v R | 5 vI |
| 1,2,3 | (7) | S | 2,6 →E |
S4. P & (Q & R), P & R → ~S, S v T |- T
We want to get T, and we have it as one disjunct of S v T, so we
should try to get a denial of the other disjunct. We have that (~S) as
the consequent of another premise (P & R → ~S), so we need
to get the antecedent of this, which is P & R. For that, we will
need to get both P and R. We have P as one conjunct of
P & (Q & R), so we can get it by &E; R is one conjunct
of the other conjunct of P & (Q & R), so we can get
it with two applications of &E.
| 1 | (1) | P & (Q & R) | |
| 2 | (2) | P & R → ~S | |
| 3 | (3) | S v T | |
| 1 | (4) | P | 1 &E |
| 1 | (5) | Q &R | 1 &E |
| 1 | (6) | R | 5 &E |
| 1 | (7) | P & R | 4,6 &I |
| 1,2 | (8) | ~S | 2,7 →E |
| 1,2,3 | (9) | T | 3,8 vE |
S5. P → Q, P → R, P |- Q & R
To get Q & R, we need to get Q, get R, and then use &I.
Both Q and R appear as consequents of conditionals that have P as
antecedent, and we already have P. So:
| 1 | (1) | P → Q | A |
| 2 | (2) | P → R | A |
| 3 | (3) | P | A |
| 1,3 | (4) | Q | 1,3 >E |
| 2,3 | (5) | R | 2,3 >E |
| 1,2,3 | (6) | Q & R | 4,5 &I |
S6. P, Q v R, ~R v S, ~Q |- P & S
To get P & S, we need to get P and S and use &I. In this
case, we already have P, so all we need to work for is S. S is a
disjunct of ~R v S, so we need to get a denial of ~R (for instance, R).
Since R appears as a disjunct of Q v R, what we need is a denial of Q:
but we have that.
| 1 | (1) | P | A |
| 2 | (2) | Q v R | A |
| 3 | (3) | ~R v S | A |
| 4 | (4) | ~Q | A |
| 2,4 | (5) | R | 2,4 vE |
| 2,3,4 | (6) | S | 3,5 vE |
| 1,2,3,4 | (7) | P &S | 1,6 &I |
S7. ~P, R v ~P ↔ P v Q, |- Q
This proof involves the rules for ↔. Basically, ↔ lets
us get a conditional from a biconditional: it can have either the left
side of the biconditional as antecendent and the right side as consequent,
or the reverse. Here, we need Q, and we have Q only as a part of one
of the parts of a biconditional (P v Q). If we could get that out,
then we could get Q using vE if we had ~P. But we're in luck, since we
do have ~P. Now all we need is a way to get P v Q out of the
biconditional. The only way to do that would be to use ↔ to
get the conditional with P v Q as consequent and then try to get its
antecedent. The conditional in question is R v ~P → P v Q. How
do we get its antecedent, R v ~P? We could do that if we had either
one of its disjuncts. Here, we can use ~P again and get R v ~P using
vI. The proof goes as follows:
| 1 | (1) | ~P | A |
| 2 | (2) | R v ~P ↔ P v Q | A |
| 1 | (3) | R v ~P | 1 vI |
| 2 | (4) | R v ~P → P v Q | 2 ↔E |
| 1,2 | (5) | P v Q | 3,4 →E |
| 1,2 | (6) | Q | 5 vE |
S8. (P ↔ Q) → R, P → Q, Q → P |- R
We have R as the consequent of a conditional, so in order to get it
we need the antecedent, P ↔ Q. Since that's a biconditional,
we need the two conditionals P → Q, Q → P in order to get it.
But those are just our second and third premises, so we're done:
| 1 | (1) | (P ↔ Q) → R | A |
| 2 | (2) | P → Q | A |
| 3 | (3) | Q → P | A |
| 2,3 | (4) | P ↔ Q | 2,3 ↔I |
| 1,2,3 | (5) | R | 1,4 →E |
S9. ~P → Q & R, ~P v S → ~T, U & ~P |- (U & R) & ~T
We've been asked to get a conjunction that has a conjunction as one of
its conjuncts. To prove this, we need to get all three of the parts: the
right conjunct ~T and the two conjuncts of the left conjunct, U and R. So,
we have three tasks. We can get U right away from the third premise by &E.
We could get ~T from the second premise if we had ~P v S, and we could get
that if we had either ~P or S, using vI. But we can get ~P from the
third premise as well, so that's solved. All that remains is getting
R. We could get it by &E from Q & R, which appears as the
consequent of the first premise, if we had the antecedent of that premise,
which is ~P. But we already needed ~P for the previous step, so that's
been solved. Here's all of this put together:
| 1 | (1) | ~P → Q & R | A |
| 2 | (2) | ~P v S → ~T | A |
| 3 | (3) | U & ~P | A |
| 3 | (4) | U | 3 &E |
| 3 | (5) | ~P | 3 &E |
| 1,3 | (6) | Q & R | 1,5 →E |
| 1,3 | (7) | R | 6 &E |
| 3 | (8) | ~P v S | 5 vI |
| 2,3 | (9) | ~T | 2,8 →E |
| 1,3 | (10) | U & R | 4,7 &I |
| 1,2,3 | (11) | (U & R) & ~T | 9,10 &I |
S10. (Q v R) & ~S → T, Q & U, ~S v ~U |- T &U
We need to get T and get U, then use &I. Getting U is easy: use &E on
Q & U. All we need to work on then is getting T. We have T as the
consequent of a conditional, (Q v R) & ~S → T, so we need to get
that conditional's antecedent, (Q v R) & ~S. Since this is a conjunction,
we need to get both conuuncts. ~S appears in ~S v ~U, so we could get it
if we had a denial of ~U, and we do (we already saw how to get U). Q v R
is a disjunction, and getting it is thus a matter of getting either of its
disjuncts and using vI. But here again, we already have U. So:
| 1 | (1) | (Q v R) & ~S → T | A |
| 2 | (2) | Q & U | A |
| 3 | (3) | ~S v ~U | A |
| 2 | (4) | Q | 2 &E |
| 2 | (5) | U | 2 &E |
| 2 | (6) | Q v R | 4 vI |
| 2,3 | (7) | ~S | 3,5 vE |
| 2,3 | (8) | (Q v R) & ~S | 6,7 &I |
| 1,2,3 | (9) | T | 1,8 →E |
| 1,2,3 | (10) | T & U | 5,9 &I |
To the Syllabus
Back to Exercise 1.4.1
On to Exercise 1.5.1
Last modified: Mon Feb 4 15:41:48 CST 2002