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Proof step 23 from proof int2bits-neg-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
~1 = 2 ^ n→i(# int2bits(~ j)) + j, j < 2 ^ (n→i(# int2bits(~ j)) - 1), 1 < 2 ^ (n→i(# int2bits(~ j)) - 1), j < 0, 0 < 2 ^ (n→i(# int2bits(~ j)) - 1) + j, 0 ≤ 2 ^ (n→i(# int2bits(~ j)) - 1), 0 < 2 ^ (n→i(# int2bits(~ j)) - 1), j + 1 < 2 ^ (n→i(# int2bits(~ j)) - 1) ⊦

Rule argument

Additional information

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