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Proof step 27 from proof byte2bits-val (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ n→i(m + n) = n→i(m) + n→i(n)
Proof step 29:
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) = 256

Rule argument

seq: ⊦ n→i(m + n) = n→i(m) + n→i(n) subst: [n, m] → [n, # int2bits(b2i(by))] rotate: T spec: intnat inst: name: n2i-add

Additional information

Used terms: n, # int2bits(b2i(by))