[我只是阅读了Lean的文档,并尝试完成3.7. Exercises,但还没有全部完成,但是这里是前四个练习(没有经典推理):
变量p q r:属性
-∧和的可交换性例如:p q = q对不起例如:p∨q↔p =:=对不起
-∧和as的关联性例如:(p q)r r p p(q r):=对不起例如:(p q)r r p p(q r):=对不起
这是我在前四个练习中所做的:
variables p q r : Prop
-- commutativity of ∧ and ∨
example : p ∧ q ↔ q ∧ p :=
iff.intro
(assume h : p ∧ q,
show q ∧ p, from and.intro (and.right h) (and.left h))
(assume h : q ∧ p,
show p ∧ q, from and.intro (and.right h) (and.left h))
example : p ∨ q ↔ q ∨ p :=
iff.intro
(assume h : p ∨ q,
show q ∨ p, from
or.elim h
(assume hp : p,
show q ∨ p, from or.intro_right q hp)
(assume hq : q,
show q ∨ p, from or.intro_left p hq))
(assume h : q ∨ p,
show p ∨ q, from
or.elim h
(assume hq : q,
show p ∨ q, from or.intro_right p hq)
(assume hp : p,
show p ∨ q, from or.intro_left q hp))
-- associativity of ∧ and ∨
example : (p ∧ q) ∧ r ↔ p ∧ (q ∧ r) :=
iff.intro
(assume h: (p ∧ q) ∧ r,
have hpq : p ∧ q, from and.elim_left h,
have hqr : q ∧ r, from and.intro (and.right hpq) (and.right h),
show p ∧ (q ∧ r), from and.intro (and.left hpq) (hqr))
(assume h: p ∧ (q ∧ r),
have hqr : q ∧ r, from and.elim_right h,
have hpq : p ∧ q, from and.intro (and.left h) (and.left hqr),
show (p ∧ q) ∧ r, from and.intro (hpq) (and.right hqr))
example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) :=
iff.intro
(assume hpqr : (p ∨ q) ∨ r,
show p ∨ (q ∨ r), from or.elim hpqr
(assume hpq : p ∨ q,
show p ∨ (q ∨ r), from or.elim hpq
(assume hp : p,
show p ∨ (q ∨ r), from or.intro_left (q ∨ r) hp)
(assume hq : q,
have hqr : q ∨ r, from or.intro_left r hq,
show p ∨ (q ∨ r), from or.intro_right p hqr))
(assume hr : r,
have hqr : q ∨ r, from or.intro_right q hr,
show p ∨ (q ∨ r), from or.intro_right p hqr))
(assume hpqr : p ∨ (q ∨ r),
show (p ∨ q) ∨ r, from or.elim hpqr
(assume hp : p,
have hpq : (p ∨ q), from or.intro_left q hp,
show (p ∨ q) ∨ r, from or.intro_left r hpq)
(assume hqr : (q ∨ r),
show (p ∨ q) ∨ r, from or.elim hqr
(assume hq : q,
have hpq : (p ∨ q), from or.intro_right p hq,
show (p ∨ q) ∨ r, from or.intro_left r hpq)
(assume hr : r,
show (p ∨ q) ∨ r, from or.intro_right (p ∨ q) hr)))
我认为这是有效的,但是很长一段时间,这是我们能做到的最好吗,或者有更好的方法在精益中编写这些证明,任何建议都将不胜感激。
如果导入精益数学的library,则策略by tauto!
应该解决所有这些问题。另外,这些都是已经存在的库定理,它们的名称类似于and_comm
。
我不认为从第一性原则可以更简短地证明这些陈述。缩短某些证明的唯一方法是删除have
和show
,并使它们的可读性降低。这是我对or_assoc
的证明,与您的证明基本相同,但没有have
和show
。
example : (p ∨ q) ∨ r ↔ p ∨ (q ∨ r) :=
iff.intro
(λ h, or.elim h (λ hpq, or.elim hpq or.inl (λ hq, or.inr (or.inl hq))) (λ hr, or.inr (or.inr hr)))
(λ h, or.elim h (λ hp, (or.inl (or.inl hp))) (λ hqr, or.elim hqr (λ hq, or.inl (or.inr hq)) or.inr))