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Proof step 4 from proof bits2uint-not-neg (previous step)

∀ bits0. # bits0 < # bits → true → (bits2uint(bits0) ≤ 0 ↔ bits2uint(bits0) = 0) ⊦ bits = [], bits2uint(bits) ≤ 0 ↔ bits2uint(bits) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bits ≠ [] ↔ (∃ bits0, boolvar. bits = bits0 + boolvar ')
Proof step 6:
∀ bits0. # bits0 < # bits → true → (bits2uint(bits0) ≤ 0 ↔ bits2uint(bits0) = 0) ⊦ ∀ bits0, boolvar. bits ≠ bits0 + boolvar ', bits2uint(bits) ≤ 0 ↔ bits2uint(bits) = 0

Rule argument

seq: ⊦ bits ≠ [] ↔ (∃ bits0, boolvar. bits = bits0 + boolvar ') subst: [bits] → [bits] rotate: F spec: list inst: bitlista name: lastdiv-rew

Additional information

Used terms: bits