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Proof step 5 from proof uint2bits-add-low (previous step)

¬ 1 < j + 2 ^ i, j < 2 ^ (i - 1), j + 1 < i, 0 < 2 ^ (i - 1), j < 0, 0 < i, 0 ≤ j + 2 ^ (i - 1), j < i, j + 1 < 2 ^ (i - 1), ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ abs(j) ≤ 2 ^ (i0 - 1) → uint2bits(j + 2 ^ i0) = false ' + fillsgn(i→n(i0), int2bits(j)) ⊦ uint2bits(j + 2 ^ i) = false ' + fillsgn(i→n(i), int2bits(j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j < 2 ^ (i - 1), j + 1 < i, 0 < 2 ^ (i - 1), j < 0, 0 < i, 0 ≤ j + 2 ^ (i - 1), j < i, j + 1 < 2 ^ (i - 1), j + 2 ^ i ≤ 1, ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ abs(j) ≤ 2 ^ (i0 - 1) → uint2bits(j + 2 ^ i0) = false ' + fillsgn(i→n(i0), int2bits(j)) ⊦ uint2bits(j + 2 ^ i) = false ' + fillsgn(i→n(i), int2bits(j))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;