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Proof step 10 from proof bits2int-ones-02 (previous step)

bits.first ⊦ bits + bits0 = [], bits = [], 2 ^ n→i(# bits0) = bits2uint(bitsinvert(bits) +1 + bits0), false ∈ bits, true ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ bits ≠ [] ∧ ¬ bits.last → bits +1 = bits.butlast + true
Proof step 12:
bitsinvert(bits) ≠ [] ∧ ¬ bitsinvert(bits).last, bits.first ⊦ bits + bits0 = [], bits = [], 2 ^ n→i(# bits0) = bits2uint((bitsinvert(bits).butlast + true) + bits0), false ∈ bits, true ∈ bits0

Rule argument

seq: ⊦ bits ≠ [] ∧ ¬ bits.last → bits +1 = bits.butlast + true subst: [bits] → [bitsinvert(bits)] rotate: F spec: bitopsaux inst: name: succ-zero-03

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = [];%bits ≠ [] ⊦ ¬ false ∈ bits → ¬ bitsinvert(bits).last; Used terms: bitsinvert(bits)