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  <description>Suc-帰納法：\[(P(0)\wedge \forall k[P(k)\rightarrow P(k+1)])\rightarrow\forall n[P(n)]\tag{1}\] 累積帰納法：\[\forall n[\forall m(1)⇒(2)：図式(1)のもと、$\forall n[\forall m(2)⇒(1)：図式(2)のもと、(ア)$P(0)$および(イ)$\forall k[P(k)\rightarrow P(k+1)]$を満たす任意の述語$P$をとり、$\forall n[P(n)]$を導く。そのために(ウ)$\forall m■</description>
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  <published>2022-12-08 11:02:31</published>
  <title>Suc-帰納法と累積帰納法</title>
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