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Proof step 3 from proof and-011x-neg (previous step)

bits = boolvar + bits2, bitsand(false ' + bits0, bits) = false ' + bits1, boolvar ⊦ bits = [], 2 ^ n→i(# bits0) = bits2uint(bits1) + bits2uint(bitsinvert(bits)) % 2 ^ n→i(# bits0) + 1, false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
bitsand(false ' + bits0, true ' + bits2) = false ' + bits1 ⊦ 2 ^ n→i(# bits0) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits0) + bits2uint(bits1), false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;