Here is a model:Domain: {1, 2, 3, 4, ...}Referents: a : 64 b : 2 c : 111Extension of S : {1, 4, 9, 16, 25, ...}Extension of P : ∅Extension of D : {h1, 2i ,h 2, 4i ,h 3, 6i ,h 4, 8i ,h 5, 10i ...}(So for every number 1, 2, 3, ... in the domain, the extension of D contains an ordered pair with that number in first position and the number twice its size in second position.)State whether the following propositions are true or false on the model, and briefly explain your answers.(i) ∀x((Sx ∧ ∃wDwx) → x = a)(ii) ∃z(z 6 = b ∧ P z) ↔ ∃x∀y(Dyx ∨ Dxc)

Which ordered pairs need to be added to the relationp = {(a,a),(a,b),(b,a),(c,c),(c,d),(d,c),(d,d)}on the set X = {a,b,c,d} to create the symmetric closure p^s of p? A (a,a) B (a,b) C (a,c) D (a,d) E (b,a) F (b,b) G (b,c) H (b,d) I (c,a) J (c,b) K (c,c) L (c,d) M (d,a) N (d,b) O (d,c) P (d,d) Q None of them

Which ordered pairs need to be added to the relationp = {(a,a),(a,d),(b,b),(b,d),(c,c),(d,a),(d,b)}on the set X = {a,b,c,d} to create the equivalence relation p* generated by p? A (a,a) B (a,b) C (a,c) D (a,d) E (b,a) F (b,b) G (b,c) H (b,d) I (c,a) J (c,b) K (c,c) L (c,d) M (d,a) N (d,b) O (d,c) P (d,d) Q None of them

Which ordered pairs need to be added to the empty relationp = {}on the set X = {a,b,c,d} to create the symmetric closure p^s of p?

Consider the following relation on set B = {a, b, {a}, {b}, {a, b}}: P = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a})}. The relation P does not satisfy trichotomy.Which ordered pairs should be included in P so that an extended relation P1 (say) would satisfy trichotomy?Choose the alternative that provides all the missing ordered pairs that should be included as elements of P1 in order for P1 to be a relation that satisfies trichotomy.(For trichotomy, each element of B must be paired with each other different element in B to form elements of P1. For example, we see that b ≠ {a} but neither (b, {a}) nor ({a}, b) are elements of P, but at least one of these elements should be an element of P1.We include (b, {a}) in P1: P1 = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a}), (b, {a}), ...}.Now check all the different elements of B and if two different elements of B are not already grouped in an ordered pair of P, they should be paired and be included as elements of P1. You can ask the question: Which ordered pairs should be included in P1 so that it will be true that forall x, y ∈  B with x ≠ y, we have (x, y) ∈  P1 or (y, x)∈  P1? Refer to study guide, p 78.)  Choose the alternative that provides all the missing ordered pairs that should be included as elements of P1 in order for P1 to be a relation that satisfies trichotomy.a.(b, {a}), (b, {b}), (b, a), ({a}, {a, b}) and ({a, b}, {b})b.({a}, b), (b, {b}), ({b}, {a, b}) and ({a, b}, {a})c.(b, {a}), ({b}, b), ({a}, {b}), ({b}, {a, b}) and ({a, b}, {a})d.(b, {a}), (b, {b}), ({a}, {b}) and ({a, b}, {a, b})

For each of the following relations defined on sets M = {5, 6, 7, 8} and N = {1, 3, 5, 7, 9},list the elements and find its domain and range.(a) S1 = {(m, n)|m + n is even}(b) S2 = {(m, n)|n = 2m − 9}(c) S3 = {(m, n)|m < n}(d) S4 = {(m, n)||m − n| ≤ 2}(e) S5 = {(m, n)|m is divisible by n}(f) S6 = {(m, n)|m + n is a prime number