leio125

\[(\mathbb{A}, +, \overline{0}, -, \cdot, \overline{1})\] \[(a+b)+c=a+(b+c) \ \ \ \ \forall a,b,c\in\mathbb{A}\\\] \[\exists \ \overline{0}:a+\overline{0}=\overline{0}+a=a \ \ \ \ \forall a\in\mathbb{A}\] \[\exists \ {-a}:a+(-a)=\overline{0} \ \ \ \ \forall a\in\mathbb{A}\] \[a+b=b+a \ \ \ \ \forall a,b\in\mathbb{A}\] \[(a\cdot b)\cdot c =a\cdot (b\cdot c) \ \ \ \ \forall a,b,c\in\mathbb{A}\] \[(a+b)\cdot c=a\cdot c+b\cdot c \ \ \ \ \forall a,b,c\in\mathbb{A}\] \[a\cdot(b+c)=a\cdot b+a\cdot c \ \ \ \ \forall a,b,c\in\mathbb{A}\] \[\exists \ \overline{1}:a\cdot\overline{1}=\overline{1}\cdot a=a \ \ \ \ \forall a\in\mathbb{A}\] \[a\cdot b=b\cdot a \ \ \ \ \forall a,b\in\mathbb{A}\] Ebbene si, qualora ci riuscissimo sarebbe tutto, troppo incoerente.. The true but unprovable statement. And if provable it would be false..