PROOFS WITH EVEN MORE RULES

In Which Theorems are Proved using Rules Derived as Well as Primitive

A theorem is simply a conclusion that can be proved from the empty set of premises. The most common technique for proving a theorem is to start with its denial as an assumption and construct a proof for it that discharges that assumption with RAA. If the theorem is a conditional, it may also be possible to prove it by assuming the antecedent, deducing the consequent, and using ->I. Here are some examples.

Exercise 1.6.1

T1 (Identity): |- P->P

1	(1)	P	A
	(2)	P->P	2 ->I (1)

This is about as short as a proof gets!

T2 (Excluded Middle): |- Pv~P

1	(1)	~(Pv~P)	A
2	(2)	P	A
2	(3)	Pv~P	2 vI
1	(4)	~P	1,3 RAA (2)
1	(5)	Pv~P	4 vI
	(6)	Pv~P	1,5 RAA (1)

T3 (Non-Contradiction): |- ~(P&~P)

1	(1)	P&~P	A
1	(2)	P	1 &E
1	(3)	~P	1 &E
	(4)	~(P&~P)	2,3 RAA (1)

T4 (Weakening): |- P->(Q->P)

1	(1)	P	A
2	(2)	Q	A
3	(3)	~P	A
1	(4)	P	1,3 RAA (3)
1	(5)	Q->P	4 ->I (2)
	(6)	P->(Q->P)	5 ->I (1)

T5 (Paradox of Material Implication): |- (P->Q)v(Q->P)

1	(1)	~((P->Q)v(Q->P))	A
2	(2)	P	A
3	(3)	Q	A
3	(4)	P->Q	3 ->I (2)
3	(5)	(P->Q)v(Q->P)	4 vI
1	(6)	~Q	1,5 RAA (3)
7	(7)	Q	A
8	(8)	~P	A
1,7	(9)	P	6,7 RAA (8)
1	(10)	Q->P	9 ->I (7)
1	(11)	(P->Q)v(Q->P)	10 vI
	(12)	(P->Q)v(Q->P)	1,11 RAA(1)

T6 (Double Negation): |- P <-> ~~P

1	(1)	P	A
2	(2)	~P	A
1	(3)	~~P	1,2 RAA (2)
	(4)	P->~~P	3 ->I (1)
	(5)	~~P->P	4 Trans
	(6)	P<->~~P	4,5 <->I

T7: |- (P<->Q)<->(Q<->P)

With derived rules, this is very easy:

1	(1)	P<->Q	A
1	(2)	Q<->P	1 <-> Comm
	(3)	(P<->Q)->(Q<->P)	2 ->I (1)
4	(4)	Q<->P	A
4	(5)	P<->Q	4 <-> Comm
	(6)	(Q<->P)->(P<->Q)	5 ->I (4)
	(7)	(P<->Q)<->(Q<->P)	3,6 <->I

T8: |- ~(P<->Q) <-> (~P<->Q)

1	(1)	~(P<->Q)	A
2	(2)	P	A
3	(3)	Q	A
3	(4)	P->Q	3 ->I(2)
2	(5)	Q->P	2 ->I(3)
2,3	(6)	P<->Q	4,5 <->I
1,3	(7)	~P	1,6 RAA(2)
1	(8)	Q->~P	7 ->I(3)
9	(9)	~P	A
10	(10)	~Q	A
3,10	(11)	P	3,10 RAA(9)
10	(12)	Q->P	11 ->I(3)
2,9	(13)	Q	2,9 RAA(10)
9	(14)	P->Q	13 ->I(2)
9,10	(15)	P<->Q	12,14 <->I
1,9	(16)	Q	1,15 RAA(10)
1	(17)	~P->Q	16 ->I(9)
1	(18)	~P<->Q	8,17 <->I
	(19)	~(P<->Q) -> (~P<->Q)	18 ->I(1)
20	(20)	~P<->Q	A
21	(21)	P<->Q	A
20	(22)	~P->Q	20 <->E
20	(23)	Q->~P	20 <->E
21	(24)	P->Q	21 <->E
21	(25)	Q->P	21 <->E
26	(26)	P	A
21,26	(27)	Q	24,26 ->E
20,21,26	(28)	~P	23,27 ->E
20,21	(29)	~P	26,28 RAA(26)
20,21	(30)	Q	22,29 ->E
20,21	(31)	P	25,30 ->E
20	(32)	~(P<->Q)	29,31 RAA(21)
	(33)	(~P<->Q) -> ~(P<->Q)	32 ->I(20)
	(34)	(~P<->Q) <-> ~(P<->Q)	19,33 <->I

T9 (Peirce's Law): |- ((P -> Q) -> P) ->P

Since this is a conditional, the straightforward approach is to assume the antecedent and try to deduce the consequent. We need RAA to finish the proof, however.

1	(1)	(P -> Q) -> P	A
2	(2)	~P	A
3	(3)	P -> Q	A
1,3	(4)	P	1,3 ->E
1,2	(5)	~(P -> Q)	2,4 RAA(3)
2	(6)	~P v Q	2 vI
7	(7)	P	A
2,7	(8)	Q	6,7 vE
2	(9)	P->Q	8 ->I(7)
1	(10)	P	5,9 RAA(2)
	(11)	((P -> Q) -> P) ->P	10 ->I(1)

T10: |- (P->Q)v(Q ->R)

We assume the denial of what we want and use RAA. With disjunctions, this is always harder than you expect. This can be shortened considerably with derived rules, obviously.

1	(1)	~((P->Q)v(Q ->R))	A
2	(2)	P->Q	A
2	(3)	(P->Q)v(Q ->R)	2 vI
1	(4)	~(P->Q)	1,3 RAA(2)
5	(5)	~P	A
6	(6)	P	A
6	(7)	P v Q	6 vI
5,6	(8)	Q	5,7 VE
5	(9)	P->Q	8 ->I(6)
1	(10)	P	4,9 RAA(5)
11	(11)	Q	A
11	(12)	P ->Q	11 ->I(6)
1	(13)	~Q	4,12 RAA(11)
1	(14)	~Q v R	13 vI
1,11	(15)	R	11,14 vE
1	(16)	Q -> R	15 ->I(11)
1	(17)	(P->Q)v(Q ->R)	16 vI
	(18)	(P->Q)v(Q ->R)	1,17 RAA(1)

T11: |- (P <-> Q) <-> (~P <-> ~Q)

1	(1)	P <-> Q	A
2	(2)	~P	A
3	(3)	~Q	A
1	(4)	P->Q	1 <->E
1	(5)	Q->P	1 <->E
6	(6)	P	A
1,6	(7)	Q	4,6 ->E
1,3	(8)	~P	3,7 RAA(6)
1	(9)	~Q ->~P	8 ->I(3)
10	(10)	Q	A
1,10	(11)	P	5,10 ->E
1,2	(12)	~Q	2,11 RAA(10)
1	(13)	~P->~Q	12 ->I(2)
1	(14)	~P<->~Q	9,13 <->I
	(15)	(P <-> Q) -> (~P <-> ~Q)	14 ->I(1)

...and that's halfway there. But the reverse direction is very similar: just interchange ~P and P, ~Q and Q in the first 15 lines. You need a final step of <->I, of course.

T12: |-

1	(1)		A
2	(2)		A
1	(3)
	(4)
	(5)
	(6)
	(7)
	(8)
	(9)
	(10)
	(11)
	(12)

T13 (& Idempotence): |- P <-> (P&P)

This is relatively simple.

1	(1)	P	A
1	(2)	P&P	1,1 &I
	(3)	P->(P&P)	2 ->I(1)
4	(4)	P&P	A
4	(5)	P	4 &E
	(6)	(P&P)->P	5 ->I(4)
	(7)	P <-> (P&P)	3,6 <->I

T14 (v Idempotence): |- P <-> (P v P)

1	(1)		A
2	(2)		A
1	(3)
	(4)
	(5)
	(6)

T15: |- (P<->Q)&(R<->S) ->((P->R)<->(Q->S))

Assume the antecedent, then get the consequent by assuming each side in turn, deducing the other side, using ->I, and then combining with <->I. The final step is then discharging the original assumption with ->I.

1	(1)	(P<->Q)&(R<->S)	A
2	(2)	P->R	A
3	(3)	Q	A
1	(4)	P<->Q	1 &E
1	(5)	R<->S	1 &E
1	(6)	P->Q	4 <->E
1	(7)	Q->P	4 <->E
1	(8)	R->S	5 <->E
1	(9)	S->R	5 <->E
1,3	(10)	P	3,7 ->E
1,2,3	(11)	R	2,10 ->E
1,2,3	(12)	S	8,11 ->E
1,2	(13)	Q->S	12 ->I(3)
1	(14)	(P->R)->(Q->S)	13 ->I(2)
15	(15)	Q->S	A
16	(16)	P	A
1,16	(17)	Q	6,16 ->E
1,15,16	(18)	S	15,17 ->E
1,15,16	(19)	R	9,18 ->E
1,15	(20)	P->R	19 ->I(16)
1	(21)	(Q->S)->(P->R)	20 ->I(15)
1	(22)	(P->R)<->(Q->S)	14,21 <->I
	(23)	(P<->Q)&(R<->S) -> ((P->R)<->(Q->S))	2 ->I(1)

T20: |- (P<->Q)->(R&P <-> R&Q)

1	(1)	P<->Q	A
2	(2)	R&P	A
1	(3)	P->Q	1 <->E
1	(4)	Q->P	1 <->E
2	(5)	R	2 &E
2	(6)	P	2 &E
1,2	(7)	Q	3,6 ->E
1,2	(8)	R&Q	5,7 &I
1	(9)	R&P -> R&Q	8 ->I(2)
10	(10)	R&Q	A
10	(11)	R	10 &E
10	(12)	Q	10 &E
1,10	(13)	P	4,12 ->E
1,10	(14)	R&P	11,13 &I
1	(15)	R&Q -> R&P	14 ->I(10)
	(16)	R&P <-> R&Q	9,15 <->I
	(17)	(P<->Q)->(R&P <-> R&Q)	16 ->I(1)

T23: |- P & (Q<->R) -> (P&Q <-> R)

1	(1)	P & (Q<->R)	A
2	(2)	P&Q	A
1	(3)	P	1 &E
1	(4)	Q<->R	1 &E
1	(5)	Q->R	4 <->E
2	(6)	Q	2 &E
1,2	(7)	R	5,6 ->E
1	(8)	P&Q -> R	7 ->I(2)
9	(9)	R	A
1	(10)	R->Q	4 <->E
1,9	(11)	Q	9,10 ->E
1,9	(12)	P&Q	3,11 &I
1	(13)	R -> P&Q	12 ->I(9)
1	(14)	P&Q <-> R	8,13 <->I
	(15)	P & (Q<->R) -> (P&Q <-> R)	14 ->I(1)

T25: |- P -> (Q ->R) <-> Q -> (P->R)

Standard strategy for a biconditional: assume each side in turn and deduce the other, use ->I twice, then use <->I. Notice that steps 9-16 are exactly like steps 1-8 with P and Q interchaged. See if you can shorten this proof by using some assumptions more than once.

1	(1)	P -> (Q ->R)	A
2	(2)	Q	A
3	(3)	P	A
1,3	(4)	Q->R	1,3 ->E
1,2,3	(5)	R	2,4 ->E
1,2	(6)	P->R	5 ->I(3)
1	(7)	Q -> (P->R)	6 ->I(2)
	(8)	(P -> (Q->R)) -> (Q -> (P->R))	7 ->I(1)
9	(9)	Q -> (P->R)	A
10	(10)	P	A
11	(11)	Q	A
9,11	(12)	P->R	9,11 ->E
9,10,11	(13)	R	10,12 ->E
9,10	(14)	Q->R	13 ->I(11)
9	(15)	P -> (Q->R)	14 ->I(10)
	(16)	(Q -> (P->R)) -> (P -> (Q->R))	15 ->I(9)
	(17)	P -> (Q ->R) <-> Q -> (P->R)	8,16 <->I

Note that on lines 8 and 16 you can't drop the parentheses around the antecedent and consequent, even though you can drop the analogous parentheses in the conclusion.

T29: |- ~P -> P <-> P

Notice that this is (~P -> P) <-> P (<-> binds more weakly than .>). This is an interesting result: what might an actual sentence of the form ~P ->P be like? The proof might be considered excessively clever.

1	(1)	~P -> P	A
2	(2)	~P	A
1,2	(3)	P	1,2 ->E
1	(4)	P	2,3 RAA(2)
	(5)	(~P->P) ->P	4 ->I(1)
6	(6)	P	A
6	(7)	~P->P	6 ->I(2)
	(8)	P->(~P->P)	7 ->I(6)
	(9)	~P -> P <-> P	5,8 <->I

T33: |- (Pv~P) & Q <-> Q

As it happens, Pv~P is a theorem (EM). We can use this in constructing a quick proof, if we jump ahead just a little (see pp. 37-38). Could you generalize this to cases involving theorems other than Pv~P?

Incidentally, the Logic Daemon doesn't know the names of theorems and will flag an error if you try to use this procedure.

1	(1)	(Pv~P) & Q	A
1	(2)	Q	1 &E
	(3)	(Pv~P) & Q -> Q	2 ->I(1)
4	(4)	Q	A
	(5)	Pv~P	EM
4	(6)	(Pv~P) & Q	4,5 &I
	(7)	Q-> (Pv~P) & Q	6 ->I(4)
	(8)	(Pv~P) & Q <-> Q	3,7 <->I

T: |-

1	(1)		A
2	(2)		A
1	(3)
	(4)
	(5)
	(6)

T8 : |-

1	(1)		A
2	(2)		A
1	(3)
	(4)
	(5)
	(6)

T39: |- (P->Q&R)->(P&Q<->P&R)

1	(1)	P->Q&R	A
2	(2)	P&Q	A
2	(3)	P	2 &E
1,2	(4)	Q&R	1,3 ->E
1,2	(5)	R	4 &E
1,2	(6)	P&R	3,5 &I
1	(7)	P&Q->P&R	6 ->I(2)
8	(8)	P&R	A
8	(9)	P	8 &E
1,8	(10)	Q&R	1,9 ->E
1,8	(11)	Q	10 &E
1,8	(12)	P&Q	9,11 &I
1	(13)	P&R->P&Q	12 ->I(8)
1	(14)	P&Q<->P&R	7,13 <->I
	(15)	(P->Q&R)->(P&Q<->P&R)	14 ->I(1)

To the Syllabus

Back to Exercise 1.5.2

On to Exercise 2.1