Question

Væ(P(x) + (Q(x) ^ S(x))) 3x(P(x) R(x)) - - .. Ex(R(x) ^ S(x)) - - -

prove that the arguments are valid using rules of inference and laws of predicate logic, (state the laws/rules used)

Answer 1

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L (Pm su tn (P()) (acul ns()). In (PC) AR (-) — R , Ix R(h)) - A In P(1) - ② and ③ Ix s(n) , (az by ) by and 8 IX (R(2) 18(M) (Proved) x (pm) Vo(r) & Gam) vs (ro) - Xx CR)73()) Inn P(2) by , tu (alms(") esince, pa by &x (s(n) TR(")) (contrapositive) since by ①, 7x TP (2), so, va P(n) can't be true. so, by , tx Q (n) - ) from A tra(n) from 5 tr s(n), since Q(u) () from 6 Vr 7R(2), since sm 7R(2) Thus, I TR(), since (Vx B(l) (7) B(n) & k o = npro L (proved)

Answer 2

a)

1)

Vr(PI) + (Q(2) AS(z)))

2)

3.0(P.1) AR(2)

From 1) we can get:-

3) [Universal instantiation using an element 'a' within the domain of x]

Pla) →Q(a) As(a)

From 2) we get:-

4) [Existential instantiation using an element 'a' within the domain of x]

Pa) A R(a)

From 4) we get:-

5) P(a) [Simplification]

From 3) and 5) we get:-

6) [Modus Ponens]

Qla) Asla

From 6) we get:-

7) S(a) [Simplification]

From 4) we get:-

8) R(a) [Simplification]

From 7) and 8) we get:-

9) [Conjunction]

R(a) A S(a)

From 9) we get:-

10) [Existential generalization]

3r(RC) A S())

Hence Proved.

b)

1)

Vr(PIVQ(2)

2)

Vr(-Q(x)V S(2))

3)

Vo(R2) +-5(2)

4) $\exists x \neg P(x)$

From 1) we get:-

5) P(a) V Q(a) [Universal instantiation using an element 'a' within the domain of x]

From 2) we get:-

6) ~Q(a) V S(a) [Universal instantiation using an element 'a' within the domain of x]

From 3) we get:-

7) R(a) -> ~S(x) [Universal instantiation using an element 'a' within the domain of x]

From 4) we get:-

8) ~P(a) [Existential instantiation using an element 'a' within the domain of x]

From 7) we get:-

9) S(a) -> ~R(a) [Contrapositive]

From 5) and 6) we get:-

10) P(a) V S(a) [Resolution]

From 10) and 8) we get:-

11) S(a) [Disjunctive syllogism]

From 9) and 11) we get:-

12) ~R(x) [Modus Ponens]

From 12) we get:-

13) [Existential generalization]

3r-R(2)

Hence Proved.

Answer 3

The solutions of the given problems as follows;

fj.: (**) (P(x) - (262) a S(x))) (b) F2.. (3x) (PGA) ARG.)- ) (widt 4.6: (1w) (RG) A S()) ^) W) Foz: P (of) (a ns): 2) Wil: of F 4 : P (a) a R (ah) .) (E) f. I off fs: ( P(a) (6) (B) Simp offy & R (a) (1) (t) An Simp of fun Fr : P (a) (Q(a)n sla)) For Q(a) a sca) () M.P of Fs and fy | Fa : Sca) () s Simp Fo fo: r R (a) a S (a) (w) da Conj. F and fa fjs (+ ) (060393 607) E: G. of Flo (CA) (x pans wer)

Fi (Wx) (P (6) v Q(x)) (2) (+b) :: F₂: (N2) (MQ(G) ~ SD 2 F: (4x) (R6) -> ~5m) i fy?' (an) (up() 9 ) (190) 9. to a: (3x) (aR(m): 3 (0 : utoto: P(Y) = Q(Y) () U.J. offi t h ~Q(y) w S(Y) () U.I. off f: R(Y) N.5(y) (1) U.I. off best to y u ~P(a) Wan (o)s F. I offe Fa: Q (a) () D.S. of Fs, Fz best to . S(a) () 21 (6) 1 D.S. of fo, fa out to : ~R. (a), (h) 3 ) ( .M.T. of Fi, Flo Fiz (1x) (wR(x)) E.G. off, s.

Answer 4

Steps NO Premise CUI) I on step o Vx (P(x) — (Q(2) ns(x))) 3x (P(x) AR (2)) . ?x (R(2) 15 (2) using rules of inference and lows of Predicate logic, he peoof is given below steps Reason x (P(x) — (8 (2) AS(2) Premise - P(a)(Q(6) Asla) Apply Universal Sustanbratisa man on Stepo - 3x (P(2) AR (2) Premise Pa) ^ R(a) for some Apply Evistential Sustantiat Ton (EI) on step P(a) for some a simplification on step Q(a) ns(a) using Modus Ponen on | Step ② and ③ R(a) for some a simplification au step & Is(a) semplification on © R(a) As(a) for some conjunction on ② and & (10 7 x (R(x) nS (2) Existential generalization on step 9 Hence Arguments are valid

Va (Pau VQ(x)) Vox ("Q(x) v S(x)) Vx (R(2) S(x)) J27 PQ Prook. ..3X-7R(W) Vel Step I steps 1 Reason 4x (P(x) v Q(x)) Premise P(a) VQ(a) Universal Lustantiation (UI) on 7x7 P(x) Premise ~P(a) for some a Existential Justantiation (EI) on ③ Q(a) Disjunctive Syllogism on and 4x(Q(2) VS (2) premise Q(a) vs (a) [UI on ② disjunctive Syllogism on 5 and 1 &&(R(x) 75(x) Premise | R(a)-75(a) TUI on UR(a) Modus Tollens on 0 and & (12) J x 7R(2) Existential Generalization on Plence arguments are valid

UI (Universal Instantiation) Rule 4x P(x) u P (c) UG (Universal Generalization) P(с) . Vxp(2) Existential Instantiation (EI) Tx P(x) P(c) for some c Existential Generalization (EG) P(c) : JxP(x) Modus Ponen Modus Tollens & q q 19 P TAP ..9 Pvq pq 278 Disjunctive Syllogism Hypothetical Syllogism TP P &

Semplification or PAQ paq 2.9 Conjunction Resolution upur Pvq png equr