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Proof step 15 from proof add-ones-lelen (previous step)

bits2 = bits0 + boolvar0 ', # bits2 ≤ # bits3, ∀ bits1, bits. # bits1 < #(boolvar + bits3) → true → ¬ false ∈ bits ∧ # bits ≤ # bits1 ∧ bits ≠ [] → bitsand(bits, bits1) = bits1 ⊦ bits3 = [], bits2 = [], bitsand(true ' + bits0, boolvar ' + bits3.butlast) + (boolvar0 ∧ bits3.last)' = boolvar ' + bits3, false ∈ bits2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# bits0 < # bits3, ∀ bits1, bits. # bits1 < #(boolvar + bits3) → true → ¬ false ∈ bits ∧ # bits ≤ # bits1 ∧ bits ≠ [] → bitsand(bits, bits1) = bits1 ⊦ bits3 = [], bitsand(true ' + bits0, boolvar ' + bits3.butlast) + bits3.last ' = boolvar ' + bits3, false ∈ bits0

Rule argument

Additional information

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