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Proof step 19 from proof int2bytes-byte (previous step)

~(2 ^ 7) + 2 ^ (6 + n→i(n0)) < 0, i + 2 ^ (6 + n→i(n0)) < 0, 0 ≤ 2 ^ (7 + n→i(n0)) + i, i < 0, ~128 ≤ i ⊦ bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) ', n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ i + ~ j < 0 ↔ i < j
Proof step 21:
2 ^ (6 + n→i(n0)) < 2 ^ 7, i + 2 ^ (6 + n→i(n0)) < 0, 0 ≤ 2 ^ (7 + n→i(n0)) + i, i < 0, ~128 ≤ i ⊦ bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) ', n0 = 0

Rule argument

spec: int-abs inst: name: neg-08 seq: ⊦ i + ~ j < 0 ↔ i < j subst: [i, j] → [2 ^ (6 + n→i(n0)), 2 ^ 7] flags: rewrite

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: 2 ^ (6 + n→i(n0)), 2 ^ 7