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Proof step 1 from proof int2uint-inv (conclusion)

⊦ j < 0 → bits2uint(bitsinvert(int2bits(j))) + 1 = ~ j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bits ≠ [] → bits2uint(bits +1) = bits2uint(bits) + 1
Proof step 3:
bitsinvert(int2bits(j)) ≠ [] ⊦ j < 0 → bits2uint(bitsinvert(int2bits(j)) +1) = ~ j

Rule argument

name: bits2uint-succ rotate: T subst: [bits] → [bitsinvert(int2bits(j))]

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ bitsinvert(bits) = [] ↔ bits = []; Used terms: bitsinvert(int2bits(j))