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Proof step 9 from proof add-ones-lelen (previous step)

1 < # bits ∧ 1 < # bits0, # 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, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# 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 = bits.first ', bits0 = bits0.first ', bits0 = [], false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ bits + boolvar = bits + boolvar ';