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Proof step 987 from proof bitsand-cast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 988:
# bits = # bits1 ⊦ bits1 = [], bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits2, bits3) + bits1)), bits3 + bits = [], bits2 + bits1 = [], bits3 = [], bits2 = [], true ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m;% ⊦ ¬ n < n;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ ¬ m ≤ n ↔ n < m;