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Proof step 11 from proof int2bytes-byte (previous step)

# int2bits(i) = n0 + 8 ∧ 8 ≤ # int2bits(i), 8 < # int2bits(i), i < 128, ~128 ≤ i ⊦ bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%lels-03: ⊦ m < n → m ≤ n; from specification nat