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Proof step 20 from proof and-011x-neg (previous step)

bits2 = bits0 + boolvar ', bitsand(false ' + bits3, true ' + bits0) + boolvar ' = false ' + bits1, ∀ 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 ⊦ bits2 = [], 2 ^ (n→i(# bits3) + 1) = 1 + bits2uint(bits1) + bits2uint(bitsinvert(bits2)) % 2 ^ (n→i(# bits3) + 1), false ∈ bits3, ¬ false ∈ bitsand(false ' + bits3, true ' + bits0) ∧ boolvar, ¬ boolvar ∈ bits1 ∧ boolvar, boolvar ∧ ¬ false ∈ bitsand(false ' + bits3, true ' + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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