P1. 1 is in N. P2. If x is in N, then its “successor” x' is in N. P3. There is no x such that x' = 1. P4. If x isn't 1, then there is a y in N such that y' = x. P5. If S is a subset of N, 1 is in S, and the implication (x in S => x' in S) holds, then S = N. Then you have to define addition recursively: Def: Let a and b be in N. If b = 1, then define a + b = a' (using P1 and P2). If b isn't 1, then let c' = b, with c in N (using P4), and define a + b = (a + c)'. Then you have to define 2:...
Hugi notar vefkökur til að bæta notendaupplifun á vefsíðunni og greina umferð um hana.
Einnig hefur Hugi uppfært persónuverndarstefnu sína. Skoðaðu stefnuna hér..
