2023年2月5日 · Specifically, for each of the $[P]$ and $[Q]$ terms in the rule (so $[Q]$ and $[R]$ in the proof) there is an extra premise(?) next to it, so it's not clear to me that disjE can be applied. How disjE is applied here?

2020年3月19日 · The rule you want to apply, namely disjunction elimination (disjE), allows you to eliminate a disjunction P ∨ Q in the premises without knowing a priori whether P or Q is true. In that situation, you need to consider both cases separately (i.e. assume P is true and assume Q is true) in order to safely eliminate the disjunction.

2019年2月9日 · In the case of r r, infer p ∧ r p ∧ r, and use ∨ Intro ∨ I n t r o to derive (p ∧ q) ∨ (p ∧ r) (p ∧ q) ∨ (p ∧ r) from that. Make sure to derive that statement in the first suproof with q q as well, and so with ∨ Elim ∨ E l i m you can then pull out (p ∧ q) ∨ (p ∧ r) (p ∧ q) ∨ (p ∧ r) You must log in to answer this question.

在Isar风格的Isabelle证明中，这很好地工作：from `a ∨ b` have fooproof assume a show foo sorrynext assume b show foo sorryqedproof在这里调用的隐式规则是rule conjE。但我应该把什么放在那里，使它不仅仅适用于一个析取：from `a ∨ b ∨ c` have fooproof(?) a

2022年2月28日 · It should be possible to do it with just those rules. assumes "(p ∨ q) ∧ (p ∨ r)" shows "p ∨ (q ∧ r)" have "p ∨ q" using assms(1) by (rule conjunct1) have "p ∨ r" using assms(1) by (rule conjunct2) moreover . {assume "p" hence "p ∨ (q ∧ r)" by (rule disjI1)} moreover . {assume "q ∧ r" hence "p ∨ (q ∧ r)" by (rule disjI2)}

$(\exists x)(p(x)\implies q)\quad \sf DisjE(1,8,17)\quad$ 2; Where at line 10, the (somewhat tedious to prove) equivalence between $\neg(\forall x)(p(x))$ and $(\exists x)(\neg p(x))$ was used to keep the proof reasonably short. The abbreviations used for the respective rules should be easily readable to anyone familiar with natural deduction.

2016年5月6日 · In Isar-style Isabelle proofs, this works nicely: assume a. show foo sorry. assume b. show foo sorry. The implicit rule called by proof here is rule conjE. But what should I put there to make it work for more than just one disjunction: proof(?) assume a. show foo sorry. assume b. show foo sorry. assume c. show foo sorry.

Information and translations of disje in the most comprehensive dictionary definitions resource on the web.

how can I prove A ⊕ B? what can I prove from A ⊕ B? . . . A n. . . . To prove A it suffices to prove A . . . 1 n. 2 . . . An. ccontr and classical are not derivable from the ND-rules. They make the …

2020年8月25日 · 直觉主义的主要原理是，通过构造性证明来建立数学命题的真。 命题联结词的意义通过证明和构造来解释： ⊥ \perp ⊥ 的证明。 这种证明解释，是布劳威尔、柯尔哥莫洛夫Kolmogorov和海廷Heyting提出的，也称为BHK解释。 imports IFOL. by (erule meta_mp) apply(erule meta_mp, assumption) done. 这叫做mp规则，中文叫做肯定前件规则。 1. (P Q) P Q . Pure.cut_rl: (PROP ?psi PROP ?theta) . PROP ?psi PROP ?theta. Pure.meta_impE: (PROP …