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बनाया गया: 8 दिसंबर 2025

का उपयोग करके उत्तर दिया गया GPT-5.1 द्वारा Chat01

प्रश्न

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Реши задачу из файла, приведи вырадение к ПНФ

विचार

Rewriting logical formula to Prenex normal form

I need to rewrite the formula ¬(P(c) ⊕ ∀x S(x)) ∨ (∀y P(y) ⇒ R(a,c)) into Prenex normal form (ПНФ). First, I'll replace the XOR and implication connectives (⊕, ⇒) with logical primitives like ¬, ∧, and ∨. Then, I'll move all quantifiers to the front, assuming c and a are constants and that P, S, and R are unary or binary predicates.

Rewriting logical formula using De Morgan’s laws

I’ll rewrite ¬(P(c) ⊕ ∀x S(x)) ∨ (∀y P(y) ⇒ R(a,c)) step by step: first, I’ll apply the XOR and implication transformations. Then, I’ll push negation using De Morgan's laws. After simplifying, I’ll replace negations of quantifiers and get to the final Prenex normal form (φ), which includes existential quantifiers.

Moving quantifiers to the front (Prenex normal form)

I’m working on rearranging the quantifiers to the front for Prenex normal form. First, I’ll handle the disjunctions and conjunctions. I need to rename some variables, like changing ∀x S(x) to ∀u S(u) and ∃x ¬S(x) to ∃v ¬S(v), to avoid conflicts between the quantifiers. Then I’ll move the quantifiers outward methodically.

Simplifying quantifier positions for Prenex form

I’ve now grouped the rightmost disjunction (∃y ¬P(y) ∨ R(a,c)) as one part C. My goal is to move the quantifiers outward, respecting some rules for disjunctions and conjunctions—such as moving ∀x or ∃x as needed while ensuring no variables are free in the wrong parts. I’ll focus on quantifier positions, since that’s what matters most for the Prenex form.

Working towards Prenex normal form

I’ve simplified part of the expression and am working on moving the quantifiers out. The goal now is to pull ∀u and ∃v outward from the conjunction inside the matrix (M1). The key is ensuring that I respect the structure of the disjunction while systematically moving those quantifiers. I need to handle this carefully while transforming M1 for Prenex.

Moving quantifiers and simplifying the structure

I clarified the steps for handling the conjunction with quantifiers, starting with pushing quantifiers outward while respecting their order of interaction. Moving quantifiers like ∃v inward was key, and I eventually ended up with φ ≡ ∃y ∃v ∀u (A(u) ∧ B(v) ∨ D(y)), where ∀u is now inside the disjunction.

Finalizing prenex form and quantifier handling

I’ve arrived at φ in prenex normal form: ∃y ∃v ∀u [ ((¬P(c) ∨ S(u)) ∧ (P(c) ∨ ¬S(v))) ∨ (¬P(y) ∨ R(a,c)) ]. The quantifiers are now at the front, though I could reorder ∃v and ∃y since they don’t depend on each other. This is equivalent, and I chose to keep it as ∃y∃v∀u for clarity.

Thought for 1m 16s

उत्तर

Формула из задания:

¬(P(c)xS(x))    (yP(y)R(a,c)).\neg\bigl(P(c)\oplus \forall x\,S(x)\bigr)\;\lor\;(\forall y\,P(y)\Rightarrow R(a,c)).

1. Убираем ⊕ и ⇒

P(c)xS(x)(P(c)¬xS(x))(¬P(c)xS(x)),P(c)\oplus \forall x S(x)\equiv \bigl(P(c)\land \neg\forall x S(x)\bigr)\,\lor\,\bigl(\neg P(c)\land \forall x S(x)\bigr), yP(y)R(a,c)¬yP(y)R(a,c).\forall y P(y)\Rightarrow R(a,c)\equiv \neg\forall y P(y)\,\lor\,R(a,c).

Получаем

¬[(P(c)¬xS(x))(¬P(c)xS(x))]    (¬yP(y)R(a,c)).\neg\Bigl[(P(c)\land \neg\forall x S(x))\lor(\neg P(c)\land \forall x S(x))\Bigr] \;\lor\;(\neg\forall y P(y)\lor R(a,c)).

2. Раскроем отрицание и заменим отрицания кванторов

(¬P(c)xS(x))(P(c)¬xS(x))    (y¬P(y)R(a,c))(¬P(c)xS(x))(P(c)x¬S(x))    (y¬P(y)R(a,c)).\begin{aligned} &(\neg P(c)\lor \forall x S(x))\land(P(c)\lor \neg\forall x S(x)) \;\lor\;(\exists y\neg P(y)\lor R(a,c))\\[2mm] &\equiv(\neg P(c)\lor \forall x S(x))\land(P(c)\lor \exists x\neg S(x)) \;\lor\;(\exists y\neg P(y)\lor R(a,c)). \end{aligned}

Переименуем связанные переменные: xS(x)uS(u)\forall x S(x)\to\forall u S(u), x¬S(x)v¬S(v)\exists x\neg S(x)\to\exists v\neg S(v):

(¬P(c)uS(u))(P(c)v¬S(v))    (y¬P(y)R(a,c)).(\neg P(c)\lor \forall u S(u))\land(P(c)\lor \exists v\neg S(v)) \;\lor\;(\exists y\neg P(y)\lor R(a,c)).

3. Вынесем кванторы наружу

¬P(c)uS(u)u(¬P(c)S(u)),P(c)v¬S(v)v(P(c)¬S(v)),y¬P(y)R(a,c)y(¬P(y)R(a,c)).\begin{aligned} \neg P(c)\lor \forall u S(u) &\equiv \forall u(\neg P(c)\lor S(u)),\\ P(c)\lor \exists v\neg S(v) &\equiv \exists v(P(c)\lor \neg S(v)),\\ \exists y\neg P(y)\lor R(a,c) &\equiv \exists y(\neg P(y)\lor R(a,c)). \end{aligned}

Тогда

[u(¬P(c)S(u))v(P(c)¬S(v))]    y(¬P(y)R(a,c)).[\forall u(\neg P(c)\lor S(u))\land \exists v(P(c)\lor \neg S(v))] \;\lor\;\exists y(\neg P(y)\lor R(a,c)).

Внутри конъюнкции:

u(¬P(c)S(u))v(P(c)¬S(v))vu[(¬P(c)S(u))(P(c)¬S(v))].\forall u(\neg P(c)\lor S(u))\land \exists v(P(c)\lor \neg S(v)) \equiv \exists v\,\forall u\bigl[(\neg P(c)\lor S(u))\land(P(c)\lor \neg S(v))\bigr].

Теперь над дизъюнкцией:

vu()    y(¬P(y)R(a,c))yvu[(¬P(c)S(u))(P(c)¬S(v))    (¬P(y)R(a,c))].\exists v\,\forall u(\dots)\;\lor\;\exists y(\neg P(y)\lor R(a,c)) \equiv \exists y\,\exists v\,\forall u\Bigl[(\neg P(c)\lor S(u))\land(P(c)\lor \neg S(v)) \;\lor\;(\neg P(y)\lor R(a,c))\Bigr].

(Порядок yv\exists y\exists v можно поменять местами, это эквивалентно.)

Ответ (ПНФ):

Например, с переименованием ux,  vzu\to x,\;v\to z:

yzx  ((¬P(c)S(x))(P(c)¬S(z))    (¬P(y)R(a,c)))\boxed{ \exists y\,\exists z\,\forall x\; \Bigl( \bigl(\neg P(c)\lor S(x)\bigr)\land\bigl(P(c)\lor \neg S(z)\bigr) \;\lor\;\bigl(\neg P(y)\lor R(a,c)\bigr) \Bigr) }

Это формула, эквивалентная исходной и записанная в пренексной нормальной форме.

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