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Proof step 4 from proof int2bits-less-2pot-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 ^ n→i(# int2bits(j)) < 2 ^ (n→i(# int2bits(j)) - 1), 0 < 2 ^ (n→i(# int2bits(j)) - 1), 2 ^ n→i(# int2bits(j)) < j, 0 ≤ j, j < 2 ^ (n→i(# int2bits(j)) - 1), 2 ^ n→i(# int2bits(j)) + 1 < 2 ^ (n→i(# int2bits(j)) - 1) ⊦

Rule argument

Additional information

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