Could someone explain to me why the following"predicate" formulas are the corresponding "english" statement.
Before the actual question. If you could answer these two side question, that would be really helpful. They are referred to as side-question below.
Please try to keep answers moderately detailed, but not too convoluted .As I am not a pro-at -logic. ——————————————————————————————————————————
Side-question 1
“All men are happy”
Correct solution, below: ∀x(M(x)→ H(x))
Incorrect ∀x(M(x) ∧ H(x)) Why is this incorrect ? ——————————————————————————————————————————
Side-question 2
“Some men are happy”
Correct solution, below: ∃x(M(x) ∧ H(x))
Incorrect ∃x(M(x)→ H(x))
Why is this incorrect ? ——————————————————————————————————————————
Actual questions below:
- All predicate statements are correct for there respective english statements.
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English statement (1) " A goal without a plan is just a dream "
Predicate statement (1) ∀x(G(x) ∧ ¬ P(x) → D(x))
Why is the above predicate statement correct for its correspond english statement ? ——————————————————————————————————————————
English statement (2) " Rainy days and Mondays always get me down "
Predicate statement (2) ∀x(R(x) v M(x) → G(x))
Why is the above predicate statement correct for its correspond english statement ? ——————————————————————————————————————————
English statement (3) “If every circuit works properly and all laptops are functioning then every test-run will terminate.”
Dictionary: C(x): x is a circuit W(x): x works properly L(x) : x is an interface F(x): x is functioning R(x): x is a test-run T(x) : x will terminate
Predicate statement (3) ∀x(C(x) → W(x)) ∧ ∀x(L(x) → F(x)) → ∀x(R(x) → T(x)).
Why isn’t (3) the following correct?
∀x∀y∀z( (C(x) ∧ W(x)) ∧ (I(y) ∧ F(y)) → (R(z) → T(z)) )
Why is the above predicate statement correct for its correspond english statement ? ——————————————————————————————————————————
Quantified predicate logic (FOL) is certainly much more technical and hard to understand compared with propositional calculus where you only have familiar truth-value connectives such as usual not, and, or, if/then. There's a critical terminology called extension. All propositional connectives are extensional such as negation: ¬A holds $\iff$ A does not; so whether ¬A holds only depends on whether A holds or doesn’t, so you can use truth table in a finite manner. But quantifiers are not extensional, for the truth value of ∃xP(x) doesn't depend on the truth value of P(x) since since P(x) contains a free variable thus it is neither true nor false. So you have to go into the semantic meaning of the quantifier, and as explained by Angus St. before you'll soon understand all your main questions.
Final note on your last question where you have a prenex normal form which you're unsure about its natural language translation correctness. Every classical first-order formula is logically equivalent to some formula in prenex normal form (not for intuitionistic logic though). There are several conversion rules that can be recursively applied to convert a formula to prenex form such as quantified DeMorgan's rule, null quantification rules, etc. Let's start your statement (3) in a Fitch like natural deduction style:
|∀x(C(x) → W(x)) ∧ ∀x(L(x) → F(x)) → ∀x(R(x) → T(x))
|∀x(C(x) → W(x)) ∧ ∀y(L(y) → F(y)) → ∀z(R(z) → T(z))
|∀x[(C(x) → W(x)) ∧ ∀y(L(y) → F(y)] → ∀z(R(z) → T(z)) ...Null Quantification on ∀x
|∀x∀y[(C(x) → W(x)) ∧ (L(y) → F(y)] → ∀z(R(z) → T(z)) ...Null Quantification on ∀y
|¬∀x∀y[(C(x) → W(x)) ∧ (L(y) → F(y)] ∨ ∀z(R(z) → T(z)) ...→ replacement
|∃x∃y[¬((C(x) → W(x)) ∧ (L(y) → F(y))] ∨ ∀z(R(z) → T(z)) ...Quantified DeMorgan's applied twice consecutively
|∀z∃x∃y[¬(((C(x) → W(x)) ∧ (L(y) → F(y))) ∨ ((R(z) → T(z)] ...Null Quantification on ∀z
|∀z∃x∃y[((C(x) → W(x)) ∧ (L(y) → F(y)) → ((R(z) → T(z))] ...→ replacement
So the correct prenex form is only slightly different than yours for the order of the 3 quantifiers and their respective types.