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Proof step 25 from proof byte2bits-val (previous step)

8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ n = 0, 2 ^ n→i(# int2bits(b2i(by))) * 2 ^ n→i(n) = 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ 0 ≤ j ∧ 0 ≤ j0 → i ^ (j + j0) = i ^ j * i ^ j0
Proof step 27:
0 ≤ n→i(# int2bits(b2i(by))) ∧ 0 ≤ n→i(n), 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ n = 0, 2 ^ (n→i(# int2bits(b2i(by))) + n→i(n)) = 256

Rule argument

seq: ⊦ 0 ≤ j ∧ 0 ≤ j0 → i ^ (j + j0) = i ^ j * i ^ j0 subst: [j0, j, i] → [n→i(n), n→i(# int2bits(b2i(by))), 2] rotate: T spec: int-pot inst: name: pot-dist-pos

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n); Used terms: n→i(n), n→i(# int2bits(b2i(by))), 2