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Proof step 84 from proof uint2bits-mod2pot (previous step)

uint2bits(i1 % 2 ^ i0) = sgn(false ' + lastn(i→n(i0), uint2bits(i1).rest)), mklist(false, # lastn(i→n(i0), uint2bits(i1).rest)) = lastn(i→n(i0), uint2bits(i1).rest), 0 < i1 % 2 ^ i0, 0 ≤ i1, 0 < i0, 0 < i1, 2 ^ i0 < i1, # uint2bits(i1) ≤ i→n(i0), ∀ j0, i. i→n(j0) < i→n(i1 * 2 + 0) → 0 ≤ i ∧ 2 ^ i < j0 → uint2bits(j0 % 2 ^ i) = sgn(false ' + lastn(i→n(i), uint2bits(j0).rest)) ⊦ i0 = 0, ~1 = i0

User applied rule insert lemma

And obtained the following new goals

Rule argument

name: uint2bits-len-min subst: [i, j] → [i0, i1]

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: i0, i1