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Proof step 6 from proof bitsand-idempotent (previous step)

¬(1 < # bits ∧ 1 < # bits), ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsand(bits0, bits0) = bits0 ⊦ bitsand(bits, bits) = bits, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits = 1, ∀ bits0. # bits0 < 1 → true → bits0 = [] ∨ bitsand(bits0, bits0) = bits0 ⊦ bitsand(bits, bits) = bits, bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);