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Proof step 16 from proof band-bitlength (previous step)

n ≤ # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(i) < # sgn(bitsand(int2bits(i), int2bits(j))), n ≤ # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(j) < n, # int2bits(i) < n, # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)) ⊦ # int2bits(i) ≤ # int2bits(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# 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)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m +1 = m + 1;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ ¬ m ≤ n ↔ n < m;