Advanced Symbolic Logic, Homework # 5 Page 74, problem # 1, part b.
p ® q. q ® r. ®. p ®r
q q. qr. ®. q r
-( q q. qr) q q r
-( q q) q -( qr) q q r
p q q q q r
Page 67 # 4 third pair
pqr q pq q pr q p pqqqr . pqqq . pqqr . pqq
pq q p p q q . p q
p p
Yes, 'p' is equivalent to 'p'
Page 74, problem # 1, part c.
p ® q. ® p : «
q q. ® p : «
-( q q) q p : «
p q p . «
p q p . .q. –( p q p )p
p q p q. –( p).p
p q p q: q q. : p
p q p q: . q q : p
p q p q: p q q p (All clauses inconsistent, so too the entire schema)
Additional problem from homework sheet:
-(pq)®: v .«
pq q: v .«
pq q : v .. q. –(v )p
pq q q q qpr
pq q q Simplified version removing obviously redundant clause
Developed Alternational Normal Form
p q r
pr q p q r
pr q p q rp q r
pr q p q rpq q rpq rq q r
pr q p q rpq q rq q r (eliminating obviously redundant clause)
Page 74, # 3
YES, this is a sound general method for testing two ANF schemata for equivalence. Each clause of an ANF schema S1 implies another ANF schema S2, if, but only if, S1 (the alternation of the implying clauses) implies S2 (which follows from (ix) p 63). If the same is true for S2 vis S1, then S2 implies S1. Since equivalence is mutual implication, this is a sound general method for testing two ANF schemata for equivalence.
Advanced Symbolic Logic
Investigate for redundant clauses and literals
pr q qr q ps q ps q s
pr ®. qr q ps q ps q s
TTT®^TqTsqT^sq^s
s
T^ pr is not redundant
qr®. pr q ps q ps q s
TT®. p^T q ps q p^s q ^s
^ q ps q ^ q ^
ps
Ts ^s
^ qr is not redundant
ps®.pr q qr q ps q s
TT®.Tr q qr q TT q T
r q qr q
patently valid ps is a redundant clause
ps ®.pr q qr q s
TTT® TT
T psis redundant
s ®.pr q qr
TT® ^ q ^ s is not redundant
pr q qr q s all redundant clauses eliminated, now test for redundant literals
r ®.pr q qr q s
TT®. pTT q ^r q^s
p p in pr is not redundant
pr .® .pr q qr q s
TT.®TT q qT q ^s
q q in pr is redundant
pr q qr q s Fully simplified
Transform into a conjunctional normal schema ( v. p. 83): (dualize, distribute, simplify, dualize)
v r v pq alternational normal form
q. qr .pqqq (dual to the original)
q r q q r . pqqq (distribution)
p q q q q rp qrq qr q p q q q q rp q rq q r
q q q rq q q rp (dropped inconsistent clauses)
q q q rp (dropped clauses that contain other clauses)
qq. q . qrqp conjunctional normal form
In the axiom system of Chapter 13, prove:
® . ®. q ® p:®: p ®. q ® p
Substitute '' for 'q' in Axiom II to get:
1. p ® . ®
Substitute ' ® ' for 'q' in Axiom I to get:
2. p ® . ® : ® :. ® . ® r : ® . p ® r
Substitute 'q ® p' for r in formula 2 to get:
3. p ® . ® : ® :. ® . ®. q ® p : ® : p ® . q ® p
Formulas 1 and 3 together with the inference rule Modus Ponens gives:
4. ® . ®.q®p:®:p®.q®p Q.E.D.