我的证明如下,但它是错误的,我不知道如何更正
assume h : ∀ x, p x ∨ r, assume a: α,
or.elim (h a)
(assume hl: p a,
show p a ∨ r, from
or.inl hl)
(assume hr: r,
show p a ∨ r, from or.inr hr)
我的证明如下,但它是错误的,我不知道如何更正
assume h : ∀ x, p x ∨ r, assume a: α,
or.elim (h a)
(assume hl: p a,
show p a ∨ r, from
or.inl hl)
(assume hr: r,
show p a ∨ r, from or.inr hr)
@lizhengxian 这个例子有帮助吗(在线精益编辑器):
variables {α : Type*} (p : α → Prop) (r : Prop) [decidable r]
example : (∀ x, p x ∨ r) ↔ (∀ x, p x) ∨ r :=
begin
split,
{ assume h,
by_cases hr : r,
{ right, assumption },
{ left,
assume x,
specialize h x,
cases h,
{ assumption },
{ contradiction } } },
{ assume h x,
cases h with h h,
{ left, apply h },
{ right, assumption } }
end