量子力學x輪迥x外星人 v0.0.1

1001 回覆
8 Like 85 Dislike
2023-09-14 01:07:26
康德有好幾個argument玩完
2023-09-14 01:08:34
U冇睇eva
2023-09-14 01:19:30
告訴我,喺非歐入面,兩條邊可唔可以構成一個圖形
2023-09-14 01:24:24
姐係一個有嘅 但冇喺domain到出現
一個原本就係冇
2023-09-14 01:25:43
哈哈,康德又少一個argument
2023-09-14 02:23:17
出現咗自己少少熟嘅term,然後又收皮
2023-09-14 22:01:12
笑乜撚你expect一個十八世紀識啲乜
2023-09-14 23:57:54
我知美國依家多數學邏輯學嘅人都係因為德國波蘭佬驚納粹走咗去啊
但點解德國波蘭佬本身數學、logic研究咁蓬勃
2023-09-15 05:56:53
Completeness theorem:
If it is true, it is provable
我估Henkin’s proof證明思路係咁:

If O is any sentence that is true, such that H |= O,
所以H union非O |= any sentence that is false,
H union with非O 自己係冇model,即係無論點樣詮釋都唔係true,即係H內derive(by universal closure)出嘅O{s} and 非O{s} 點樣詮釋都唔係true

By Henkin proof,但仲未睇
If what a set of sentence prove is consistent, then it has a model such that these are true in that model(已進行割除量化詞手術)
換句話講
If a set of sentence does not have any model, what these set of sentence proves is not consistent

所以what H union O proves is not consistent,
換句話講H union with 非O |- any sentence that is false, as our deduction system support proof by contradiction, so it is iff H |- O,
So if O is true, O is provable
Since O is arbitrary, H|= O, H|-O

所以H|= O,則H|-O
2023-09-15 14:41:00

如果所有subset啲sentence就有,咁原本嗰set就要有
咁就證埋Compactness theorem,咁勁嘅leon henkin
2023-09-15 16:21:56

唔通又係物理勿理
2023-09-15 17:18:26
When life gives you lemon, make it a lemonade
2023-09-15 19:06:28
是不是我錯覺 點解佢同tarski咁似樣
2023-09-15 20:51:29
洗唔洗
2023-09-15 22:19:20
咁多都被小圓救咗,得佢冇
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