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Proof step 981 from proof bitsand-cast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 982:
# bits = # bits1 ⊦ bits1 = lastn(# bits1, bits1), bits3 = [], bits1 = [], bits3 + bits = [], true ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m + n < m;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ fillsgn(# bits, bits) = bits;