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Proof step 11 from proof uint2bits-2potrec (previous step)

i→n(i + 1) < # uint2bits(j), 2 ^ i < j, 0 ≤ i, 2 ^ i ≤ j ⊦ uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + fillsgn(i→n(i) + 1, sgn(false ' + lastn(i→n(i), uint2bits(j).rest))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i→n(i) + 1 < # uint2bits(j), 2 ^ i ≤ j, 0 ≤ i, 2 ^ i < j ⊦ uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + fillsgn(i→n(i) + 1, sgn(false ' + lastn(i→n(i), uint2bits(j).rest))).rest

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i +1 = i + 1;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ ¬ 2 ^ i < 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;