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Proof step 21 from proof int2bytes-byte (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0

Rule argument

spec: int-pot inst: name: ls-pos-2pot-01 seq: 0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0 subst: [j0, j] → [7, 6 + n→i(n0)] flags: rewrite

Additional information

Used simplifier rules: numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%num-n2i-ls-07: i < 0 ⊦ (* i) ≤ n→i(n); from specification intnat%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%nat2int-14: ⊦ n→i(n) < 1 ↔ n = 0; from specification intnat Used terms: 7, 6 + n→i(n0)