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Proof step 29 from proof and-011x-neg (previous step)

2 ^ n→i(# bits3) = 1 + bits2uint(bitsinvert(bits0)) % 2 ^ n→i(# bits3) + bits2uint(bits4) ∨ false ∈ bits3, bitsand(false ' + bits3, true ' + bits0) = false ' + bits4, ∀ bits, bits2, bits1. # bits < #(bits3 + true ') → bitsand(false ' + bits, true ' + bits2) = false ' + bits1 → 2 ^ n→i(# bits) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits) + bits2uint(bits1) ∨ false ∈ bits ⊦ 2 ^ (n→i(# bits3) + 1) = 1 + bits2uint(bits4 + boolvar ') + bits2uint(bitsinvert(bits0) + (¬ boolvar)') % 2 ^ (n→i(# bits3) + 1), false ∈ bits3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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