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Proof step 2 from proof bits2int-ones-02 (previous step)

⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + bits2int(bits + bits0) = 0, false ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bits.first ∧ bits ≠ [] → bits2int(bits) = ~ bits2uint(bitsinvert(bits) +1)
Proof step 4:
¬((bits + bits0).first ∧ bits + bits0 ≠ []) ⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + bits2int(bits + bits0) = 0, false ∈ bits
Proof step 6:
(bits + bits0).first ∧ bits + bits0 ≠ [] ⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + ~ bits2uint(bitsinvert(bits + bits0) +1) = 0, false ∈ bits

Rule argument

name: bits2int-neg-03 rotate: F subst: [bits] → [bits + bits0]

Additional information

Used terms: bits + bits0