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Proof step 25 from proof int2bits-neg-03 (previous step)

n→i(# int2bits(~ j)) = i + 1, ~1 = 2 ^ n→i(# int2bits(~ j)) + j, j < 2 ^ i, 1 < 2 ^ i, j < 0, 0 < 2 ^ i + j, 0 ≤ 2 ^ i, 0 < 2 ^ i, j + 1 < 2 ^ i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
~1 = 2 ^ (i + 1) + j, n→i(# int2bits(~ j)) = i + 1, 0 ≤ 2 ^ i, j + 1 < 2 ^ i, 0 < 2 ^ i + j, j < 0, j < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < i + j ↔ 0 < j;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i +1 = i + 1;% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)