共同構造的幸福
2023-9-15 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
共同構造的幸福
2023-9-15 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
潘朵拉的盒拓撲
2023-9-15 15:37:05
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潘朵拉的盒拓撲
2023-9-15 15:42:53
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共同構造的幸福
2023-9-15 16:21:56
:^(
:^(
唔通又係物理勿理
共同構造的幸福
2023-9-15 17:18:26
When life gives you lemon, make it a lemonade