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Proof step 6 from proof bxor-bitlength (previous step)

# int2bits(i) < n, # int2bits(j) < n, n ≤ # int2bits(bor(band(i, bcompl(j)), band(bcompl(i), j))), # int2bits(i) < # int2bits(bor(band(i, bcompl(j)), band(bcompl(i), j))), # int2bits(j) < # int2bits(bor(band(i, bcompl(j)), band(bcompl(i), j))) ⊦ # int2bits(band(i, bcompl(j))) < n ∧ # int2bits(band(bcompl(i), j)) < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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