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Proof step 20 from proof byte2bits-val (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [2 ^ n→i(n), 1, i] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: 2 ^ n→i(n), 1, i