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Proof step 18 from proof and-011x-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];