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4) Use truth tables to verify equalities.a. pv(qvr) ≡ (pvq)vrpq(pvq)r(pvq)vr(qvr)pv(qvr)TTTTTTTTTTFTTTTFTTTTTTFTFTFTFTTTTTTFTTFTTTFFFTTTTFFFFFFFb. p^(q^r) ≡ (p^q)^rpq(p^q)r(p^q)^r(q^r)p^(q^r)TTTTTTTTTTFFFFTFFTFFFTFFFFFFFTFTFTFFTFFFFFFFFTFFFFFFFFFF8) Use deMorgan’s laws to negate the following: -(p ^v q) ≡ -p v^ -qa.(Kwame will take a job in industry) OR (will go to graduate school).(Kwame will NOT take a job in industry) AND (will NOT go to graduate school).b.(Yoshiko knows Java) AND (knows calculus).(Yoshiko does NOT know Java) OR (does NOT know calculus).c.(James is young) AND (is strong).(James is NOT young) OR (is NOT strong).d.(Rita will move to Oregon) OR (move to Washington).(Rita will NOT move to Oregon) AND (NOT move to Washington).12)Prove each statement is a tautology by using equivalencies:a.[-p^(pvq)] -> q[(-p^p) v (-p^q)] -> q[ F v (-p^q)] -> qT v -(-p^q) v qTb.[(p->q)^(q->r)] ->(p->r)(p->q) ->(p->r) v (q->r) ->(p->r)(-p v q) ->(-p v r) v (-q v r) ->(-p v r)(p ^-q) v (-p v r) v (q ^ -r) v (-p v r)(p ^-q) v (q ^ -r) v (-p v r)(p ^-q) v (q v (-p v r) ^ (-r v (-p v r))(p ^-q) v (q v (-p v r) ^ T(p ^-q) v (q v -p v r)(p v q v -p v r) ^ (-q v q v -p v r)T ^ TTc.[p^(p->q)]->q[p^(-p v q)]->q[( p^-p) v (p^q)]->q[F v (p^q)]->q(p^q)->q(-pv-q)vqTd.[(pvq) ^ (p->r) ^ (q->r)]->r[(pvq) ^ (-pvr) ^ (-qvr)]->r(-p ^ -q) v (p ^ -r) v [(q ^ -r) v r](-p ^ -q) v (p ^ -r) v [ (q v r) ^ (-r v r)](-p ^ -q) v (p^-r) v [ (q v r) ^ T](-p ^ -q) v (p^-r) v (q v r)[(-p ^ -q) v q] v [ (p^-r) v r)][(-p v q) ^ (-q v q)] v [ (p v r) ^ (-r v r)][(-p v q) ^ T] v [ (p v r) ^ T](-p v q) v (p v r)(-p v p) v q v rT v q v rT14) Is: (-p^(p->q)) -> -q :a tautology? NO, it is a “contingency”.p-pq(p->q)(-p^(p->q))-q(-p ^ (p->q))->-qTFTTFFTTFFFFTTFTTTTFFFTFTTTT18) Show that: (p->q) ≡ (-q -> -p).By table:pq-q-p(p -> q)(-q -> -p)TTFFTTTFTFFFFTFTTTFFTTTTBy “rules”:p -> q-p v qq v –p-q -> -p26) Show that: -p -> (q -> r) ≡ q -> (p v r). By table:pqr-p(q -> r)-p -> (q -> r)(p v r)q -> (p v r)TTTFTTTTTTFFFTTTTFTFTTTTTFFFTTTTFTTTTTTTFTFTFFFFFFTTTTTTFFFTTTFTBy “rules”:-p -> (q -> r)p v (-q v r)-q v (p v r) q -> (p v r)32) Show that: (p ^ q) -> r) not ≡ (p -> r) ^ (q -> r). By table:pqr(p ^ q)(p -> r)(q -> r)(p ^ q) -> r(p -> q) ^ (q -> r)TTTTTTTTTTFTFFFFTFTFTTTTTFFFFTTFFTTFTTTTFTFFTFTFFFTFTTTTFFFFTTTT ................
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