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Proof step 36 from proof int2bits-len-neg (previous step)

¬(0 ≤ i ∧ 2 ^ i ≤ ~ j ∧ ~ j < 2 ^ (i + 1)), true ∈ uint2bits(~ j).rest.rest, j < 2 ^ i, j < ~1, j + 2 ^ i < i, j + 2 ^ i < 0, 0 ≤ 2 ^ i * 2 + j, 0 ≤ i, 0 < 2 ^ i, j < i, j + 1 < 2 ^ i, j < 0 ⊦ # uint2bits(~ j) = i→n(i + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
0 = 2 ^ i * 2 + j, 0 ≤ i, 0 < 2 ^ i, j + 2 ^ i < 0, j < 0, j + 1 < 2 ^ i, j < i, j + 2 ^ i < i, j < ~1, j < 2 ^ i, true ∈ uint2bits(~ j).rest.rest, 2 ^ i + j < 0 ⊦ # uint2bits(~ j) = i→n(i + 2)

Rule argument

Additional information

Used simplifier rules: ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ (* i1) * j + i < j ↔ (i1 - 1) * j + i < 0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ 0 ≤ i ∧ i ≤ 1 → i = 0 ∨ i = 1; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ i * j = j * i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i < j → ¬ j < i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;