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Proof step 11 from proof byte2bits-val (previous step)

8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ bits2uint(mklist(true, n) + int2bits(b2i(by))) = b2i(by) + 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ # bits0 ≠ 0 → bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0)
Proof step 13:
# int2bits(b2i(by)) ≠ 0, 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ bits2uint(mklist(true, n)) * 2 ^ n→i(# int2bits(b2i(by))) + bits2uint(int2bits(b2i(by))) = b2i(by) + 256

Rule argument

seq: ⊦ # bits0 ≠ 0 → bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0) subst: [bits0, bits] → [int2bits(b2i(by)), mklist(true, n)] rotate: F spec: int2bits inst: name: bits2uint-split

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ []; Used terms: int2bits(b2i(by)), mklist(true, n)