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Proof step 4 from proof bitsand-zeros-02 (previous step)

¬(¬ true ∈ bits ∧ (# bits ≤ # bits0 ∨ bits0.first) ∧ bits ≠ [] ∧ bits0 ≠ []) ⊦ bits = [], bits0 = [], sgn(bitsand(bits, bits0)) = false ', true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# bits0 < # bits ⊦ sgn(bitsand(bits, bits0)) = false ', bits0 = [], bits = [], bits0.first, true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;