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Proof step 7 from proof add-ones-lelen (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# bits ≤ 1 + # bits2, ∀ bits1, bits. # bits1 < #(boolvar + bits2) → true → ¬ false ∈ bits ∧ # bits ≤ # bits1 ∧ bits ≠ [] → bitsand(bits, bits1) = bits1 ⊦ bits = [], bitsand(bits, boolvar ' + bits2) = boolvar ' + bits2, false ∈ bits, bits ≠ bits.first ' ∧ bits2 ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;