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Proof step 1 from proof and-one-01 (conclusion)

bits ≠ [], bits0 ≠ [] ⊦ bitsand(bits, bits0) = true ' + bits1 → bits.first ∧ bits0.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bitsand(bits, bits0) = true ' + bits1, bits0.first, true ∈ bitsand(bits, bits0) ⊦ bits0 = [], bits = [], bits.first

Rule argument

Additional information

Used simplifier rules: and-notone-03: ⊦ bits ≠ [] ∧ bits0 ≠ [] ∧ ¬ bits0.first → bitsand(bits, bits0) ≠ true ' + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;