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Proof step 111 from proof uint2bits-restsgn (previous step)

uint2bits(2 * 2 ^ i - i0 + 1) = boolvar ' + bits, 0 < 2 ^ i, 1 < 2 ^ i, i0 < 2 ^ i, 2 ^ i ≤ i0 * 2, 0 < i, 0 ≤ 2 ^ i, i0 * 2 + 1 < 4 * 2 ^ i, 0 ≤ i, 0 ≤ i0, 0 < i0, i0 < 2 * 2 ^ i, ∀ j0, i. i→n(j0) < i→n(i0 * 2 + 1) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ boolvar, 4 * 2 ^ i = 2 * (i0 + 1), sgn(bits + true ') = bits + true ', false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 112:
uint2bits(2 * 2 ^ i - i0 + 1) = false ' + bits, 0 < 2 ^ i, 1 < 2 ^ i, i0 < 2 ^ i, 2 ^ i ≤ i0 * 2, 0 < i, 0 ≤ 2 ^ i, i0 * 2 + 1 < 4 * 2 ^ i, 0 ≤ i, 0 ≤ i0, 0 < i0, i0 < 2 * 2 ^ i, ∀ j0, i. i→n(j0) < i→n(i0 * 2 + 1) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ 4 * 2 ^ i = 2 * (i0 + 1), sgn(bits + true ') = bits + true ', false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ boolvar ∈ bits → (boolvar ∈ boolvar0 ' + bits ↔ boolvar ↔ boolvar0);% ⊦ i * j = j * i;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)