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Proof step 6 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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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)) ⊦

Rule argument

Additional information

Used simplifier rules: and-notempty-09: ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)