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Proof step 13 from proof and-011x-neg (previous step)

false ' + bits0 ≠ [], false ' + bits0 = false ' + bits1, ∀ bits, bits2, bits1. # bits < # bits0 → 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(# bits0) = 1 + bits2uint(bits1), false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ bits, bits2, bits3. # bits < # bits1 → bitsand(false ' + bits, true ' + bits2) = false ' + bits3 → 2 ^ n→i(# bits) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits) + bits2uint(bits3) ∨ false ∈ bits ⊦ 2 ^ n→i(# bits1) = 1 + bits2uint(bits1), false ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;