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Proof step 6 from proof int2bits-less-2pot-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n→i(# int2bits(j)) = i + 1, 0 < 2 ^ i, 2 ^ (i + 1) < 2 ^ i, 2 ^ (i + 1) + 1 < 2 ^ i, j < 2 ^ i, 0 ≤ j, 2 ^ (i + 1) < j ⊦

Rule argument

Additional information

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