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Proof step 19 from proof band-bitlength (previous step)

¬(¬ int2bits(j).first ∧ # int2bits(j) ≤ # int2bits(i) ∧ int2bits(j) ≠ []), # int2bits(j) < n, # int2bits(j) < # int2bits(i), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) < n, n ≤ # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(i) < # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(j) < # sgn(bitsand(int2bits(i), int2bits(j))), n ≤ # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < n, # int2bits(j) + 1 < # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ m < n → ¬ n < m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j).first ↔ j < 0;