Prove the following predicate logic statements valid or invalid: A) ∀x(A(x) → B(x)) ∧ ∀x(A(x) ∨ ¬C(x)) ∧ ∃x(¬B(x)) → ¬∃x(¬A(x) ∧ ¬B(x) ∧ ¬C(x)) B) ∀x(A(x) → B(x)) ∧ ∃x(A(x) ∨ B(x)) → ∃x(A(x) ∧ B(x))
Prove the following predicate logic statements valid or invalid:
A) ∀x(A(x) → B(x)) ∧ ∀x(A(x) ∨ ¬C(x)) ∧ ∃x(¬B(x)) → ¬∃x(¬A(x) ∧ ¬B(x) ∧ ¬C(x))
B) ∀x(A(x) → B(x)) ∧ ∃x(A(x) ∨ B(x)) → ∃x(A(x) ∧ B(x))
A) To prove the validity of the statement, we assume that the antecedent is true and the consequent is false, and show that this leads to a contradiction.
Assume that the antecedent is true and the consequent is false. This means that:
- ∀x(A(x) → B(x)) is true: for all values of x, if A(x) is true, then B(x) is true.
- ∀x(A(x) ∨ ¬C(x)) is true: for all values of x, either A(x) is true or ¬C(x) is true (or both).
- ∃x(¬B(x)) is true: there exists at least one value of x for which ¬B(x) is true.
- ¬∃x(¬A(x) ∧ ¬B(x) ∧ ¬C(x)) is false: there does not exist a value of x for which ¬A(x) ∧ ¬B(x) ∧ ¬C(x) is true.
From ∃x(¬B(x)) we know that there exists at least one value of x, say a, for which ¬B(a) is true. Using this and the first premise, we can conclude that A(a) → B(a) is true. Since ¬B(a) is true, it follows that A(a) must be false.
Using the second premise, we can conclude that either A(a) is true or ¬C(a) is true. Since A(a) is false, it must be the case that ¬C(a) is true.
Now consider the expression ¬A(a) ∧ ¬B(a) ∧ ¬C(a). Since ¬B(a) and ¬C(a) are true, it follows that ¬A(a) must also be true. This means that ¬A(a) ∧ ¬B(a) ∧ ¬C(a) is true for the value of x = a, which contradicts the assumption that ¬∃x(¬A(x) ∧ ¬B(x) ∧ ¬C(x)) is false.
Therefore, we have a contradiction and our initial assumption that the antecedent is true and the consequent is false must be false. This means that the statement is valid.
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