难以理解逻辑
好吧,所以我必须证明以下序列:
(p -> r) ^ (q -> r) |- p ^ q -> r
我理解为什么这显然是正确的,并且我也理解自然演绎的规则。 我不明白的是我如何去证明它。 以下是提供的标准答案:
1. (p -> r) ^ (q -> r) |- p ^ q -> r premise
2. p ^ q assumption
3. p ^e 2
4. p -> r ^e 1
5. r ->e 4,3
6. p ^ q -> r ->i 2,5
(e = elimination / i = introduction).
有人可以向我提供链接或“简单”的解释吗? 我觉得我错过了一个简单的概念,导致这很难理解......?
例如,在第 4 行,为什么需要第 3 行中的 p 来删除 ->,而在第 3 行中,您可以在不使用 aq 的情况下删除 ^ q ?
我确信这很简单,但对我来说似乎没有意义......?
Ok so I have to prove the following sequent:
(p -> r) ^ (q -> r) |- p ^ q -> r
I understand why that is clearly correct and I also understand the rules of natural deduction. What I don't understand is how I go about proving it. Here is the model answer provided:
1. (p -> r) ^ (q -> r) |- p ^ q -> r premise
2. p ^ q assumption
3. p ^e 2
4. p -> r ^e 1
5. r ->e 4,3
6. p ^ q -> r ->i 2,5
(e = elimination / i = introduction).
Could someone provide me with a link or a 'dumbed-down' explanation? I feel like I am missing a simple concept that is causing this to be hard to understand... ?
For example, on line 4, why does it require the p from line 3 to remove the ->, where as in line 3, you can remove the ^ q without using a q?
I am sure this is quite straight forward but it doesn't seem to make sense to me... ?
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您可以在不使用 q 的情况下删除 ^ q,因为 p ^ q 表示 p AND q - p 是独立于 q 的。
您无法删除 p -> 不使用 p 因为 p -> r 意味着 p 隐含 r——只有当 p 也是 true 时,r 才保证为真。
You can remove the ^ q without using q because p ^ q means p AND q -- p is true independent of q.
You can't remove the p -> without using p because p -> r means p IMPLIES r -- r is only guaranteed to be true if p is as well.
在第 2 行中,您有
p ^ q
,这意味着p
和q
都为 true。 由此可见,p
为真,因为如果两者都为真,则任何一个也为真。在第 4 行中,仅当
p
为 true 时,r
才为 true。 在第 3 行中,p
为 true。 因此,r
也是正确的。In line 2, you have
p ^ q
which means that bothp
andq
are true. From that follows thatp
is true, because if both of them are true, then any single one is also true.In line 4,
r
is true only ifp
is true. And in line 3 you have thatp
is true. Therefore,r
is also true.