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Proof step 13 from proof shiftzero-by-one-01 (previous step)

i < 0 ∧ i ≠ ~1, i < ~1, i < 0 ⊦ fillsgn(n - 1, int2bits((i - 1) / 2)) = [], n = 0, 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = 2 ^ n→i(# fillsgn(n, int2bits((i - 1) / 2) + (~1 = i % 2)') - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i < ~1, i < 0 ⊦ i = ~1, fillsgn(n - 1, int2bits((i - 1) / 2)) = [], n = 0, 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = 2 ^ n→i(# fillsgn(n, int2bits((i - 1) / 2) + (i % 2 ≠ 0)') - 1)

Rule argument

Additional information

Used simplifier rules: ⊦ j < 0 → (j % 2 ≠ ~1 ↔ j % 2 = 0);