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Proof step 10 from proof cast-min (previous step)

¬(0 < i ∧ i ≠ 1), cast(i, j) < 0, i→n(i - 1) + 1 < # int2bits(cast(i, j)), cast(i, j) < ~1, 0 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
cast(1, j) < ~1, cast(1, j) < 0 ⊦ # int2bits(cast(1, j)) = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i = j → i - j = 0;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;