PROOFS WITH EVEN FEWER TEARS

Being the Continuation of a Student's Handy Guide to Proving Sequents in the system of Allen/Hand, Logic Primer

Some more exercises from Allen/Hand, Section 1.5.1

S29. ~(P&Q) -||- ~Pv~Q

Left to right: No point repeating the argument to show that you can't get this one by vI, even though that is the first strategy to think of with disjunctions. (In this case, it just won't work). Instead, we have to resort to the indirect approach, which turns out not to be as bad as you might expect. Assume ~(~Pv~Q), then look for a contradictory pair. One nice contradictory pair to look for would be P&Q and ~(P&Q), since we already have ~(P&Q). In order to get P&Q, we need to get P and to get Q. The proof below does each of those indirectly: assume ~P (which at once gives you ~Pv~Q, the denial of (2)), then discharge with RAA; then do the analogous thing with ~Q. This gives P&Q, which is the denial of the premise, and so we can use RAA to discharge the assumption at (2).
1(1)~(P&Q) A
2(2)~(~Pv~Q)A
3 (3)~PA
3(4)~Pv~Q 3 vI
2(5)P2,4 RAA (3)
6(6)~QA
6(7)~Pv~Q6 vI
2 (8)Q2,7 RAA (6)
2(9)P&Q 5,8 &I
1(10)~Pv~Q1,9 RAA (2)

Moral: it's often a good idea to try the indirect approach when proving disjunctions.

Right to left:
1(1)~Pv~Q A
2(2)P&QA
2 (3)P2 &E
2(4)Q2 &E
1,2(5) ~Q1,3 vE
1 (6)~(P&Q)4,5 RAA (2)

S27. P -> Q, ~P -> Q |- Q

Your text names this 'Special Dilemma', but another medieval name for it was 'Consequentia Mirabilis': 'the amazing consequence.'

This requires a little subtlety. To get Q out of (1), we'd need P, and there's no likely way to get it. So, best to try the indirect approach: we assume ~Q, then try to get a contradiction. Here is where it becomes subtle. There are two likely contradictory pairs we could aim for: Q and ~Q or P and ~P. Since we've assumed ~Q, it might be easiest to try to get Q. We could get Q if we had either P (from (1)) or ~P (from (2)). So, we could try to get one of those. I will choose ~P and try to get it with the indirect approach: assume P and look for a contradictory pair. In fact, we can get Q right away, giving us a contradictory pair with (3); ~P then gives us Q, so we have a contradictory pair after discharging the assumption of P, and we can then discharge the first assumption (~Q):
1(1)P->Q A (premise)
2(2)~P->QA (premise)
3(3)~QA
4(4)PA
1,4(5)Q1,4 ->E
1,3(6)~P3,5 RAA (4)
1,2,3(7)Q2,6 ->E
1,2(8)Q3,7 RAA (3)

Notice that we can't stop at line (7), since its assumption set still includes the undischarged assumption (3). However, since we have contradictory lines (namely, it and that very assumption), we can discharge (3) with RAA.

S24. P -> Q -||- ~(P & ~Q)

Left to right: use the indirect approach. After you assume P&~Q, it's relatively quick to get both Q and ~Q, then use RAA. Right to left: what you want is a conditional, so use ->I (assume P and try to get Q). There's no rules for getting at the parts of negated conjunctions, so assume ~Q and look for a contradictory pair (the easiest one to get is ~(P&~Q) and P&~Q).
Left to right:
1(1)P->QA (premise)
2(2)P&~QA
2 (3)P2 &E
1,2(4)Q1,3 ->E
2(5) ~Q2 &E
1(6)~(P&~Q)4,5 RAA (2)
Right to left:
1(1)~(P&~Q)A (premise)
2(2) PA
3(3)~QA
2,3(4)P&~Q2,3 &I
1.2(5)Q1,4 RAA (3)
1(6)P->Q 5 ->I (2)

S36. P & Q -||- Q & P

This is easy, but it's important to realize that it does require proof. Here's the left-to-right part:
1(1)P&Q A (premise)
1(2)P1 &E
1(3)Q1 &E
1(4) Q&P2,3 &I

Right-to-left bears an extraordinary similarity to this.

S37. P v Q -||- Q v P

Compare this with the previous proof. It's not quite as easy. The only choice here is the indirect approach. After assuming ~(QvP), we need to get a contradictory pair. We can aim for the denial of that very assumption, that is, the thing we want to prove: QvP. To get that, we'd need either P or Q. However, we can get either one of those from (1) if we have the denial of the other. So, we assume one of them (in this case, I chose ~P):
1(1)PvQ A (premise)
2(2)~(QvP)A
3(3)~PA
1,3(4)Q1,3 vE
1,3 (5)QvP4 vI
1,2(6)P2,5 RAA (3)
1,2(7) QvP6 vI
1 (8)QvP2,7 RAA (2)

S51. ~(P<->Q) -||- P<->~Q

Strategy here is important. We'll begin with the left-to-right proof. Since what we want to conclude is a biconditional, we will need to use <->I after first deducing both P->~Q and ~Q->P. We also know that we need to get each of these with "1" (the number of our first assumption) as the entire assumption set, since that's what we want on the last line. So, we know this much:
1(1) ~(P<->Q)A (premise)
....
1(m) P->~Q
....
1(n) ~Q->P
1( )P<->~Qm,n <->I
Since each of the things we need to deduce is a conditional, the strategy we should use in each case is to assume the antecedent and try to get the consequent, then use ;->I. We can therefore fill in a little more of our proof:
1(1) ~(P<->Q)A (premise)
2(2)PA
....
(m-1)~Q
1(m)P->~Q->I (2)
m+1(m+1)~QA
....
(n-1)P
1(n) ~Q->P->I (m+1)
1(n+1)P<->~Qm,n <->I
Now all we need to do is fill in the proof between (2) and (m-1) and the proof between (m+1) and (n-1). The first of these is going to require a use of RAA, since we need to get ~Q and don't currently have it as part of anything. So, we will make yet another assumption, this time of the denial of ~Q (i.e. Q), and try to get a contradiction. So, our proof will look like this:
1(1) ~(P<->Q)A (premise)
2(2)PA
3(3)QA
....
(m-1)~Q ?, ?, RAA(3)
1(m)P->~Q->I (2)
m+1(m+1)~QA
....
(n-1)P
1(n) ~Q->P->I (m+1)
1(n+1)P<->~Qm,n <->I
How will we then get a contradiction? At this point, we need to think about the first premise, which we haven't made any use of at all. This is a negation, so if we could manage to get P<->Q, we would have a contradiction. To get that, in turn, we would need to get P->Q and Q->P and use <->I:
1(1) ~(P<->Q)A (premise)
2(2)PA
3(3)QA
....
(m-4)P->Q
(m-3)Q->P
(m-2)P<->Qm-3, m-4 <->I
(m-1)~Q 1,m-2 RAA(3)
1(m)P->~Q->I (2)
m+1(m+1)~QA
....
(n-1)P
1(n) ~Q->P->I (m+1)
1(n+1)P<->~Qm,n <->I
So, all that remains is to get lines (m-4) and (m-3). These are conditionals, so in each case we will assume the antecedent and try to get the consequent. But notice that we have, in each case, already assumed the antecedent, and we've already got the consequent (as an assumption, in each case. So, we can do this:
1(1) ~(P<->Q)A (premise)
2(2)PA
3(3)QA
3(4)P->Q3 ->I (2)
2(5)Q->P2 ->I (3)
2,3(6)P<->Q4, 5 <->I
1,2(7)~Q 1,6 RAA(3)
1(8)P->~Q->I (2)
9(9)~QA
....
(n-1)P
1(n) ~Q->P->I (m+1)
1(n+1)P<->~Qm,n <->I
Now we're halfway done. All we have to do is the reverse. Sort of. Let's try to use the same strategy as before. We'll try to get P at line (n-1) by assuming ~P and deducing a contradiction, and as before we'll try to get that contradiction by first getting P-Q and Q-P and then deducing P-Q, the denial of (1):
1(1) ~(P<->Q)A (premise)
2(2)PA
3(3)QA
3(4)P->Q3 ->I (2)
2(5)Q->P2 ->I (3)
2,3(6)P<->Q4, 5 <->I
1,2(7)~Q 1,6 RAA(3)
1(8)P->~Q->I (2)
9(9)~QA
10(10)~PA
....
(n-4)P->Q
(n-3)Q->P
(n-2)P<->Qn-4,n-3 <->I
(n-1)P1, n-2, RAA (10)
1(n) ~Q->P->I (9)
1(n+1)P<->~Qm,n <->I
Now, we might think we could just do what we did before in lines (4) and (5). However, we need to remember that we need the right assumption sets in front of the lines. At line (n), we're using ->I to get rid of assumption number (9), so it's a good bet that line (n-1) should have 9 in its assumption set. For a similar reason, line (n-2) is probably going to have (10) in its assumption set. And if (n-1) has 9 in its assumption set, then the only place it could get it would be from (n-2), so we can assume it has a 9 as well:
1(1) ~(P<->Q)A (premise)
2(2)PA
3(3)QA
3(4)P->Q3 ->I (2)
2(5)Q->P2 ->I (3)
2,3(6)P<->Q4, 5 <->I
1,2(7)~Q 1,6 RAA(3)
1(8)P->~Q->I (2)
9(9)~QA
10(10)~PA
....
(n-4)P->Q
(n-3)Q->P
9,10(n-2)P<->Qn-4,n-3 <->I
9(n-1)P1, n-2, RAA (10)
1(n) ~Q->P->I (9)
1(n+1)P<->~Qm,n <->I
These are clues to how we will need to fill in the rest of the proof. The problem with just repeating lines (4) and (3) to get our contradiction is that we get the wrong assumption set: we would have 2,3 instead of 9,10. What we need, then, is a way of getting P that doesn't have a new assumption number attached. There's a way to do this. We have both P (line (2)) and ~P (line (10)), so we can use this contradiction to infer the denial of any previous assumption: and ~Q (at (9)) is an assumption, so we can use RAA to get Q as the denial of line (9). Since the assumption we are discharging is (9), that gives (2,10) as the assumption set; we then get rid of 2 by using ->>I to deduce line (n-4). Changing this a little, we use the contradictory pair (9) and (3) to get P by RAA, then infer line (n-3) by ->I. This gives us the finished proof, so we can now fill in all the actual line numbers:
1(1) ~(P<->Q)A (premise)
2(2)PA
3(3)QA
3(4)P->Q3 ->I (2)
2(5)Q->P2 ->I (3)
2,3(6)P<->Q4, 5 <->I
1,2(7)~Q 1,6 RAA(3)
1(8)P->~Q->I (2)
9(9)~QA
10(10)~PA
2,10(11)Q 2,10 RAA (9)
3,9(12)P 3,9 RAA (10)
10(13)P->Q11 ->I (2)
9(14)Q->P12 ->I (3)
9,10(15)P<->Q13,14 <->I
9(16)P1, 15, RAA (10)
1(17) ~Q->P16 ->I (9)
1(18)P<->~Q8,17 <->I
Astonishingly, the whole thing actually works out.

There's a lot of other exercises in here you could do...

Some exercises from Allen/Hand, Section 1.5.2

S54. P-> Q & R, R v ~Q -> S & T, T <-> U |- P -> U

Overall strategy: assume P, get U, use ->I. U occurs on the right side of a biconditional, so we can get it if we can get the other side (use <->E, then ->E), which is T. To get T, we need to get the right side of (2), for which we need its left side Rv~Q. Since that's a disjunction, we can get it if we can get either of its disjuncts R and ~Q. We can get R from the right side of (1) by &E; to get that, we'll need P. But we've already assumed P, so we've found the proof:
1(1)P->Q&R A (premise)
2(2)Rv~Q->S&TA (premise)
3(3)T<->UA (premise)
4(4)PA
1,4(5)Q&R1,4 ->E
1,4(6)R5 &E
1,4(7) Rv~Q6 vI
1,2,4 (8)S&T2,7 ->E
1,2,4(9)T8 &E
3(10) T->U3 <->E
1,2,3,4(11)U9,10 ->E
1,2,3(12)P->U11 ->I (4)

S55. (~P v Q) & R, Q -> S |- P -> (R -> S)

The conclusion is a conditional, so assume P and try to deduce R->S. Since that's also a conditional, assume R and try to get S. To get S we need Q, and we can get that from the left side of (1) and vE if we can get P; but we've already assumed P. So, we can discharge the two assumptions of R and P, in that order, to get the conclusion.
1(1)(~PvQ)&R A (premise)
2(2)Q->SA (premise)
3(3)PA
4(4)RA
1(5)~PvQ1 &E
1,3(6) Q3,5 vE
1,2,3 (7)S2,6 ->E
1,2,3(8)R->S 7 ->I (4)
1,2(9)P->(R->S)8 ->I (3)

Notice that this is just a mite sneaky. We didn't actually use the assumption (4) to get anything at all: we got S without using it, then discharged it to get R->S. Sneaky, but legitimate.

S56. Q & R, Q -> P v S, ~(S & R) |- P

Use the indirect approach. Having assumed P, what shall we try to contradict? Since it's hard to think of anything else to do with ~(S&R), let's try for that. We can get its denial, S&R, if we can get both S and R. R is easy (line (1) by &E), so we're halfway there. We could get S if we could get the right side of (2) and use vE (we already have assumed ~P). But Q comes from (1) by &E, so we're home:
1(1)Q&R A (premise)
2(2)Q->PvSA (premise)
3(3)~(S&R)A (premise)
4(4)~PA
1(5)Q1 &E
1,2 (6)PvS2,5 ->E
1,2,4(7)S4,6 vE
1(8) R1 &E
1,2,4(9)S&R7,8 &I
1,2,3(10)P3,9 RAA (4)

S57. P -> R & Q, S -> ~R v ~Q |- S & P -> T

1(1)P->R&Q A (premise)
2(2)S->~Rv~QA (premise)
3(3)S&PA
4(4)~TA
3(5)S3 &E
2,3(6) ~Rv~Q2,5 ->E
3(7)P3 &E
1,3(8) R&Q1,7 ->E
1,3(9)R8 &E
1,2,3(10)~Q6,9 vE
1,3(11)Q8 &E
1,2,3(12)T10,11 RAA (4)
1,2(13)S&P->T 12 ->I (3)

S58. R & P, R -> (S v Q), ~(Q & P) |- S

S is a disjunct of the right side of (2), so if we can get that disjunct out (using ->E, requires R) we can get S with vE (this will require ~Q). R we can get from (1) with &E. ~Q is less obvious; we'll need to use the indirect approach. However, notice that if we assume Q and can get P, we can get Q&P, which is the denial of (3): so, there's our contradictory pair. Here's the proof:
1(1)R & PA (premise)
2(2)R -> (S v Q)A (premise)
3(3)~(Q & P)A (premise)
4(4)QA
1(5)P1 &E
1,4 (6)Q&P4,5 &I
1,3(7)~Q3,6 RAA (4)
1(8)R1 &E
1,2(9)SvQ2,8 ->E
1,2,3(10)S7,9 vE

S59. P & Q, R & ~S, Q -> (P -> T), T -> (R -> S v W) |- W

We could get W using vE, if we could get SvW out of line (4) and get ~S. ~S is available from (2) by &E. However, to get to SvW, we need T and R. R comes from line (2); T would come from the right side of (3), if we had P and Q. But we can get both those from (1), so we're set.
1(1)P & QA (premise)
2(2)R & ~SA (premise)
3(3)Q -> (P -> T)A (premise)
4(4)T -> (R -> S v W)A (premise)
1(5)P1 &E
1(6)Q1 &E
1,3(7) P->T3,6 ->E
1,3(8)T5,7 ->E
1,3,4(9)R->SvW4,8 ->E
2(10)R2 &E
1,2,3,4(11)SvW9,10 ->E
2(12)~S2 &E
1,2,3,4(13)W11,12 vE

S60. R -> ~P, Q, Q -> (P v ~S) |- S -> ~R

Main strategy: Assume S and try to get ~R, then use ->E. To get ~R, assume R and aim for a contradictory pair, then use RAA. There's an S and a ~S in the premises, so the contradictory pair to try for is S and ~S. We already have S, so it's just a matter of getting ~S. To get that, we need first to get the right side of (3) , for which we need Q; but we already have Q. Then, we need ~P, for which we need R; but we've already assumed that. So, we can do the proof:
1(1)R -> ~PA
2(2)QA
3(3)Q -> (P v ~S)A
4(4)SA
5(5)RA
1,5(6)~P1,5 ->E
2,3(7)Pv~S2,3 ->E
1,2,3,5(8)~S6,7 vE
1,2,3,4(9)~R4,8 RAA (5)
1,2,3(10)S->~R9, ->I (4)

S61. P -> Q, P -> R, P -> S, T -> (U -> (~V -> ~S)), Q -> T, R -> (W -> U), V -> ~W, W |- ~P

This looks hideous, but we can approach it systematically. We want to assume P and aim or a contradictory pair, then use RAA. We have several possibilities hidden in the premises: w and ~W, S and ~S, V and ~V. Since W all by itself is a premise, let's try to get ~W. For that, we need V. We can get V (maybe) if we can get out the right side of the right side of (4), then use ->E twice (for which we will need T and U). We can get T from (5) if we can get Q, and we can get Q from (1) if we have P (but we've assumed it at (9). As for U, we can get that if we can get R and W; W we already have, and R we can get from (2) if we have P (and once again, we've assumed it at (9)). So, we can get ~V->~S. What we needed was V; so, we assume ~V and look for a contradictory pair, which is easy to find (get ~S from ~V->~S, S from (3) and (9)). Or, in its full glory:
1(1)P -> QA (premise)
2(2)P -> RA (premise)
3(3)P -> SA (premise)
4(4)T -> (U -> (~V -> ~S))A (premise)
5 (5)Q -> TA (premise)
6(6)R -> (W -> U)A (premise)
7(7)V -> ~WA (premise)
8(8)WA (premise)
9(9)P A
1,9(10)Q1,9 ->E
2,9 (11)R2,9 ->E
3,9(12)S3,9 ->E
1,5,9(13)T5,10 ->E
1,4,5,9(14)U -> (~V -> ~S)4,13 ->E
2,6,9(15)W -> U6,11 ->E
2,6,8,9(16)U8,15 ->E
1,2,4,5,6,8,9(17)~V -> ~S14,16 ->E
18(18)VA
7,18(19)~W7,18 ->E
7,8(20)~V8,19 RAA (18)
1,2,4,5,6,7,8,9(21)~S17,20 ->E
1,2,3,4,5,6,7,8(22)~P12,21 RAA (9)

S62. P <-> ~Q & S, P & (~T -> ~S) |- ~Q & T

We need to build a conjunction, so we need to get each of its conjuncts. ~Q is available by &E if we can get the right hand side of (1), so we'll want to use <->E and then look for P to use with the resulting conditional and ->E. To get T, we need first to get at the right hand side of P (use &E), then try assuming T for an RAA. The latter is relatively straightforward: we get ~S from ~T->~S, then get S from ~Q&S. Here (after a little tidying up) is the proof:
1(1)P <-> ~Q & SA (premise)
2(2)P & (~T -> ~S)A (premise)
1(3)P->~Q&S1, <->E
2(4)P2 &E
1,2(5) ~Q&S3,4 ->E
1,2(6)S5 &E
2(7) ~T->~S2 &E
8(8)~TA
2,8(9)~S7,8 ->E
1,2(10)T6,9 RAA (8)
1,2(11)~Q5 &E
1,2(12)~Q & T10,11 &I

S63. P v Q <-> P & Q -||- P <-> Q

This is lengthy but relatively straightforward. Since each side is a biconditional, each of the two proofs will consist of two subproofs, each of which is the proof of a conditional. For left-to-right, we assume P and get Q, then assume Q and get P, then get the biconditional by <->I. In each case, we need only the left-right conditional from the premise.
1(1)P v Q <-> P & QA (premise)
1(2)PvQ -> P&Q 1 <->E
3(3)PA
3(4)PvQ3 vI
1,3(5) P&Q2,4 ->E
1,3(6)Q5 &E
1(7) P->Q6 ->I (3)
8(8)QA
8(9)PvQ8 vI
1,8 (10)P&Q2,9 ->E
1,8(11)P10 &E
1(12) Q->P11 ->I (8)
1(13)P<->Q7,12 <->I
1(1)P<->Q A (premise)
1(2)P->Q1 <->E
1(3)Q->P 1 <->E
4(4)PvQA
5(5)~PA
4,5(6)Q4,5 vE
1,4,5 (7)P3,6 ->E
1,4(8)P5,7 RAA (5)
1,4(9)Q2,8 ->E
1,4(10)P&Q 8,9 &I
1(11)PvQ->P&Q10 ->I (4)
12(12)P&QA
12 (13)P12 &E
12(14)PvQ13 vI
(15) P&Q->PvQ14 ->I (12)
1(16)PvQ<->P&Q 11,15 <->I

Notice that the assumption set for line (15) is empty. That's not an error: line 15 is a theorem which can be proved from no assumptions at all.