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Proof step 978 from proof bitsand-cast (previous step)

¬(# bits1 = # bits ∧ bits2 ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ []), # bits = # bits1 ⊦ bits3 = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits2 + bits1, bits3 + bits))), bits3 + bits = [], bits2 + bits1 = [], bits1 = [], true ∈ bits1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 979:
# bits = # bits1 ⊦ bits3 = [], bits1 = [], bits3 + bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits1, bits3 + bits))), true ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;