"also have...finally have" 在伊莎贝尔中的用法
Usage of "also have...finally have" in Isabelle
我通常认为 also have
是这样工作的:
have "P r Q1" by simp
also have "... r Q2" by simp
also have "... r Q3" by simp
...
also have "... r Qn" by simp
finally have "P r Qn+1" by simp
其中 "... r Qm"
表示 "Qm-1 r Qm"
而 r
是某种传递关系。
r
的意思是 =
这似乎确实如此,但是,当使用 ≥
时,我发现了这个描述的反例:
...
have "1- 1/(2^(n+1))≥1/(2::real)" by simp
also have "... ≥ 0" (* here when I check the 'output' it seems to be
considering "0 ≤ 1 - 1 / 2 ^ (n + 1)" which in the previous notation
would be Qn r Qn+2 !*)
我的问题是,also have
是如何工作的,特别是,我如何预测 ...
将指代什么?
AFAIR,a ≥ b
是 b ≤ a
的缩写。考虑到这一点,您会发现这不符合 also
.
预期的模式
我建议你用另一种方式陈述你的不等式链,从最低到最高。如果你愿意,你仍然可以使用 ≥
来表示最终结果——毕竟它只是一个缩写。
我通常认为 also have
是这样工作的:
have "P r Q1" by simp
also have "... r Q2" by simp
also have "... r Q3" by simp
...
also have "... r Qn" by simp
finally have "P r Qn+1" by simp
其中 "... r Qm"
表示 "Qm-1 r Qm"
而 r
是某种传递关系。
r
的意思是 =
这似乎确实如此,但是,当使用 ≥
时,我发现了这个描述的反例:
...
have "1- 1/(2^(n+1))≥1/(2::real)" by simp
also have "... ≥ 0" (* here when I check the 'output' it seems to be
considering "0 ≤ 1 - 1 / 2 ^ (n + 1)" which in the previous notation
would be Qn r Qn+2 !*)
我的问题是,also have
是如何工作的,特别是,我如何预测 ...
将指代什么?
AFAIR,a ≥ b
是 b ≤ a
的缩写。考虑到这一点,您会发现这不符合 also
.
我建议你用另一种方式陈述你的不等式链,从最低到最高。如果你愿意,你仍然可以使用 ≥
来表示最终结果——毕竟它只是一个缩写。