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Proof step 9 from proof bitsand-idempotent (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsand(bits0, bits0) = bits0 ⊦ bits = [], bitsand(bits.butlast, bits.butlast) + bits.last ' = bits, # bits = 1

Rule argument

Additional information

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