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Proof step 1002 from proof bitsand-cast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1003:
# bits0 ≤ # bits1, bits0.first ⊦ bits2 + bits1 = [], bits1 = lastn(# bits1, fillsgn(# bits1, bits2 + bits1)), bits0 = [], bits1 = [], true ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n ≤ m ↔ m = n);% ⊦ ¬ true ∈ bits → mklist(false, # bits) = bits;