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Proof step 10 from proof band-bitlength (previous step)

int2bits(i) ≠ [] ∧ int2bits(j) ≠ [], # int2bits(i) ≤ # int2bits(j), n ≤ # bitsand(int2bits(j), int2bits(i)), # int2bits(j) < # sgn(bitsand(int2bits(j), int2bits(i))), # int2bits(i) < # sgn(bitsand(int2bits(j), int2bits(i))), n ≤ # sgn(bitsand(int2bits(j), int2bits(i))), # int2bits(j) < n, # int2bits(i) < n, # int2bits(j) < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < # bitsand(int2bits(j), int2bits(i)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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