(i) If the compound proposition “(p → q) ∧ (p ∧ r)” is false

1.	(i) If the compound proposition “(p → q) ∧ (p ∧ r)” is false, then find the truth values of p, q and r. (ii) If the compound proposition p→ (q ∨ r) is false, then find the truth values of p, q and r. (iii) If the compound proposition p → (~q ∨ r) is false, then find the truth values of p, q and r. (iv) If the truth value of the propositions (p ∧ q) → (r ∨ ~s) is false, then find the truth values of p, q, rand s.
Answer» (i) Given (p → q) ∧ (p ∧ r) is false (a) Case 1: p → q is true & p ∧ r is false p is T q is T & p is T &ris F p = T, q = T, r = F Case 2(a): p = F, q = T p ∧ r= F → p = F, r = F p = F, q = T, r = f (b): (p →q) is F & par is true T → F = F T ∧ T is T P=T, q = F, r=T Case 3: (p → q) is F & (p ∧ r) is false T → F = F F ∧ F = F F ∧ T = F F ∧ F = F . ∴ p = T, q = F, r= F. (ii) Given p → (q ∨ r) is false T → F = F ∴ p = T & q ∨ r is false = F ∨ F= F ∴ p = T, q = F & r= F (iii) Given p → (q ∨ r) is false Then T → F= F ∴ P = T, ~ q ∨ r= F F ∨ F = F ∴ q = T, q = T, r = F. (iv) Given (p ^ q) → (r ∨ ~s) is false We know that T → F = F ∴ p∧q = T and r ∨ ~s = F is false T∧ T = T F ∨ F= F is false ∴ p = T, q = T, r = F, S = T

(i) If the compound proposition “(p → q) ∧ (p ∧ r)” is false, then find the truth values of p, q and r. (ii) If the compound proposition p→ (q ∨ r) is false, then find the truth values of p, q and r. (iii) If the compound proposition p → (~q ∨ r) is false, then find the truth values of p, q and r. (iv) If the truth value of the propositions (p ∧ q) → (r ∨ ~s) is false, then find the truth values of p, q, rand s.

Answer»

(i) Given (p → q) ∧ (p ∧ r) is false

(a) Case 1: p → q is true & p ∧ r is false

p is T q is T & p is T &ris F

p = T, q = T, r = F

Case 2(a): p = F, q = T p ∧ r= F → p = F, r = F

p = F, q = T, r = f

(b): (p →q) is F & par is true

T → F = F T ∧ T is T

P=T, q = F, r=T

Case 3: (p → q) is F & (p ∧ r) is false

T → F = F

F ∧ F = F

F ∧ T = F

F ∧ F = F .

∴ p = T, q = F, r= F.

(ii) Given p → (q ∨ r) is false

T → F = F

∴ p = T & q ∨ r is false

= F ∨ F= F

∴ p = T, q = F & r= F

(iii) Given p → (q ∨ r) is false

Then T → F= F

∴ P = T, ~ q ∨ r= F

F ∨ F = F

∴ q = T, q = T, r = F.

(iv) Given (p ^ q) → (r ∨ ~s) is false

We know that T → F = F

∴ p∧q = T and r ∨ ~s = F is false

T∧ T = T

F ∨ F= F is false

∴ p = T, q = T, r = F, S = T