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Proof step 1019 from proof bitsand-cast (previous step)

¬ true ∈ bits1 ∧ (# bits1 ≤ # bits0 ∨ bits0.first) ∧ bits1 ≠ [] ∧ bits0 ≠ [], # bits0 = # bits1 ⊦ bits1 = [], bits2 + bits1 = [], bits1 = lastn(# bits1, fillsgn(# bits1, bits1)), bits0 = [], true ∈ bits1, bits0.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1020:
# bits0 = # bits1 ⊦ bits1 = lastn(# bits1, bits1), bits2 + bits1 = [], bits1 = [], bits0 = [], true ∈ bits1, bits0.first

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m;% ⊦ fillsgn(# bits, bits) = bits;