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Proof step 21 from proof byte2bits-val (previous step)

2 ^ n→i(n) = i + 1, 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ bits2uint(int2bits(b2i(by))) + 2 ^ n→i(# int2bits(b2i(by))) * i = b2i(by) + 256, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ j < 0 → bits2uint(int2bits(j)) = 2 ^ n→i(# int2bits(j)) + j
Proof step 23:
b2i(by) < 0, 2 ^ n→i(n) = i + 1, 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ (2 ^ n→i(# int2bits(b2i(by))) + b2i(by)) + 2 ^ n→i(# int2bits(b2i(by))) * i = b2i(by) + 256, n = 0

Rule argument

seq: ⊦ j < 0 → bits2uint(int2bits(j)) = 2 ^ n→i(# int2bits(j)) + j subst: [j] → [b2i(by)] rotate: F spec: int2bits inst: name: int2uint-neg

Additional information

Used terms: b2i(by)