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Proof step 2 from proof uint2bits-falsein (previous step)

1 < i, 0 < i, 0 ≤ i ⊦ false ∈ bitsinvert(uint2bits(i)) +1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bits ≠ [] → (false ∈ bits +1 ↔ ¬(¬ bits.last ∧ ¬ false ∈ bits.butlast))
Proof step 4:
bitsinvert(uint2bits(i)) ≠ [], 1 < i, 0 < i, 0 ≤ i ⊦ bitsinvert(uint2bits(i)).last ∨ false ∈ bitsinvert(uint2bits(i)).butlast

Rule argument

seq: ⊦ bits ≠ [] → (false ∈ bits +1 ↔ ¬(¬ bits.last ∧ ¬ false ∈ bits.butlast)) subst: [bits] → [bitsinvert(uint2bits(i))] rotate: F spec: bitopsaux inst: name: succ-falsein-02

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ bitsinvert(bits) = [] ↔ bits = []; Used terms: bitsinvert(uint2bits(i))