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Questions tagged [predicate-logic]

Questions concerning predicate calculus, i.e. the logic of quantifiers.

Some well-known formal systems covered by this term are

  • first-order logic, containing the quantifiers $\forall$ and $\exists$
  • second-order logic
  • many-sorted logic
  • infinitary logic
4144 questions
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Predicate for a simple statement

Today is my first day working with predicate logic. Here is what is given (I've written it down myself not sure if this is the correct way): G(w) → w is a genre L(x, G(w)) → x likes w genre F(x, y) → x is from country y S(m,g) → Suggest m mood song…
Chaudhary Talha
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Saturation of infinite trees

What is (are) the difference(s) between saturated and unsaturated trees in predicate logic? We have two trees: Tree A: $\forall x \exists y (Fx \land Gy) \backslash ab$ $\exists y (Fa \land Gy)\checkmark b$ $(Fa \land…
Hanlon
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write the following **sentences** into **quantified statements.**

In a domain of natural numbers, use $P(x)$ for $x$ is prime and $Q(x)$ for $x$ is even. Also the symbols $ \ <,>$ for $x$ and $=$ for $y$ can be used. Then write the following sentences into quantified statements. (i) Some primes are even. (ii)…
MAS
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Prove the statement $\forall x\in\mathbb N (x > 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$.

Can someone please show me how to prove this statement? $$\forall x\in\mathbb N (x > 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$$ I can only assume $m = 3$ and when $x > 1$.
jhg
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Negating a quantified formula NOR

I want to negate the following sentence: All redheaded persons in the room have a pet , that either cat or dog is I must not negate the quantifier.
S Hristoskov
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What does this symbol mean (set theory)

I hate when I come across symbols I can't recognise or describe because it's nigh-on impossible to google for them, so I hope it's okay to post this here. Does anyone know what this symbol means within the realm of Set Theory/Predicate Logic: In…
Leon Aves
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