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Proof step 11 from proof add-ones-lelen (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# bits2 < # bits0, ∀ bits1, bits. # bits1 < # bits0 → true → ¬ false ∈ bits ∧ # bits ≤ # bits1 ∧ bits ≠ [] → bitsand(bits, bits1) = bits1 ⊦ bitsand(true ' + bits2.butlast, bits0.butlast) + (bits2.last ∧ bits0.last)' = bits0, bits0 = [], bits0 = bits0.first ', bits2 = [], false ∈ bits2

Rule argument

Additional information

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