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Proof step 18 from proof and-ass (previous step)

¬(bits0 ≠ [] ∧ bits1.first ' = true '), ∀ bits2, bits0, bits1. # bits2 < # bits → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bits = [], bits0 = [], bitsand(bits, bitsand(bits0, bits1.first ')) = bitsand(bitsand(bits, bits0), bits1.first ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ bits2, bits0, bits1. # bits2 < # bits → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bits = [], bits0 = [], false ' = bitsand(bitsand(bits, bits0), false '), bits1.first

Rule argument

Additional information

Used simplifier rules: and-zero-03: ⊦ bits ≠ [] → bitsand(bits, false ') = false ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;