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Proof step 14 from proof uint2bits-2pot-01 (previous step)

0 < 2 ^ (i1 + 1), 0 ≤ j, 0 ≤ i1, 0 < j, ~1 ≤ i1, ∀ i0, j. i→n(i0) < i→n(i1 + 1) → 0 ≤ i0 ∧ 0 < j → uint2bits(j * 2 ^ i0) = uint2bits(j) + mklist(false, i→n(i0)) ⊦ j * 2 ^ (i1 + 1) = 1, uint2bits(j * 2 ^ i1) + false ' = uint2bits(j) + false ' + mklist(false, i→n(i1))

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0, j] → [i1, j]

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);% ⊦ 0 ≤ i → abs(i) = i;% ⊦ ¬ i < i;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: i1, j