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Proof step 11 from proof prime-divides-pot-01 (previous step)

prime(i), i | i0 ^ i1, 0 < i1, ∀ j0, i, i0. i→n(j0) < i→n(i1 + 1) → prime(i) ∧ 0 < j0 → (i | i0 ^ j0 ↔ i | i0) ⊦ i1 = 0, i | i0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i, i0] → [i1, i, i0]

Additional information

Used simplifier rules: abs-succ-03: ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1; from specification intnat%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i1, i, i0