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Proof step 6 from proof int2bytes-byte (previous step)

¬ # int2bits(i) ≤ 8, ~128 ≤ i, i < 128 ⊦ bits2bytes(int2bits(i)) = i2b(i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
8 < # int2bits(i), i < 128, ~128 ≤ i ⊦ bits2bytes(int2bits(i)) = i2b(i) '

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ []; from specification int2bits%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat