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