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Proof step 14 from proof int2bytes-byte (previous step)

¬(0 ≤ i ∧ 1 < n0 + 8), # int2bits(i) = n0 + 8, i < 128, ~128 ≤ i ⊦ n0 = 0, 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: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%numlslsdel: j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0); from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%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%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1