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Proof step 4 from proof band-bitlength (previous step)

# int2bits(i) < n, # int2bits(j) < n, n ≤ # int2bits(bits2int(bitsand(int2bits(i), int2bits(j)))), # int2bits(i) < # int2bits(bits2int(bitsand(int2bits(i), int2bits(j)))), # int2bits(j) < # int2bits(bits2int(bitsand(int2bits(i), int2bits(j)))) ⊦

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 ≠ 0; (forward)%bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);