This is an assignment of introduction to logic on coursera.
Consider a language object constants a and b and relation constants p and q where p has arity 1 and q has arity 2. The following is a truth assignment for this language.
p(a) = 1
p(b) = 0
q(a,a) = 0
q(a,b) = 1
q(b,a) = 1
q(b,b) = 0
Say whether each of the following sentences is true or false under with this truth assignment.
I can't solve the evaluation of this two sentences:
∀x.p(x) ⇒ ∃y.q(y,y)
Here is how I figure,
For ∀x.p(x) ⇒ ∃y.q(y,y). To close the sentence, I got ∀x.∃y.(p(x) ⇒ q(y,y)).Instances of this sentence is a) ∃y.(p(a) ⇒ q(y,y)), b) ∃y.(p(b) ⇒ q(y,y)).If both instances is true according to the truth assignment, then ∀x.p(x) ⇒ ∃y.q(y,y) is true.
First to determine the truth of ∃y.(p(a) ⇒ q(y,y)), find all instances of p(a) ⇒ q(y,y),there are two possibilities,1) p(a) ⇒ q(a,a),2) p(a) ⇒ q(b,b),looking at the truth assignment,p(a) ⇒ q(a,a) is false,p(a) ⇒ q(b,b) is false.Since both these instances is false, the existential sentence ∃y.(p(a) ⇒ q(y,y)) as a whole is false.
Since ∃y.(p(a) ⇒ q(y,y)) is false according to the truth assignment, ∀x.∃y.(p(x) ⇒ q(y,y)) is false, finally ∀x.p(x) ⇒ ∃y.q(y,y) is false, but the answer is TRUE.
Thanks.
Best Answer
$\forall x.p(x)$ is false because $p(b)$ is false. An implication with false antecedent is true, so the sentence is true.