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Proof step 16 from proof uint2bits-sgn-2pot (previous step)

0 ≤ 2 ^ n→i(# bits), 0 < 2 ^ n→i(# bits), ∀ i. ¬ i < 0 → 2 ^ n→i(# bits) ≠ 2 ^ i ⊦ sgn(true ' + bits) = true ' + true ' + bits, true ∈ bits

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n→i(# bits)

Additional information

Used simplifier rules: ⊦ ¬ n→i(n) < 0; Used terms: n→i(# bits)