Let P(x), Q(x), R(x) and S(x) denote the following predicates with domain Z: P(x): x ≀ 0, Q(x): x2 = 1, R(x): x is oddS(x): x = x + 1.For each predicate, determine its truth value

Accepted Solution

Answer: Hi!, first, Z are the integer numbers, so we only will work with them.P(x): x ≀ 0ok, this predicate is true if x is less or equal tan 0, and false if x is greater than 0.so P(x) is true if { x∈Z, x ≀ 0} Q(x): x2 = 1Q(x) is true only if 2*x = 1. now, this means that if x=1/2 is true, but 1/2 isnt an integer, then Q(x) is false βˆ€ x ∈ Z.R(x): x is oddR(x) is true if x is odd, we can write odd numbers as x = 2k + 1, where k is a random integer; then:R(x) is true if x=2k +1, with k∈Z.S(x): x = x + 1S(x) is true if x= x+1, if we subtract x from both sides of the equality, we get that S(x) is true if 0=1, and this is absurd, then:S(x) is false Β βˆ€ x ∈ Z.