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Proof step 20 from proof bitsor-ones-false (previous step)

∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ # bits = 1, bits1 = [], bits = [], bitsor(bits.butlast, false ' + bits1.butlast) + (bits.last ∨ bits1.last)' = bits, false ∈ bits

User applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
bits = bits2 + boolvar ', ∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ # bits = 1, bits1 = [], bits = [], bitsor(bits2, false ' + bits1.butlast) + (boolvar ∨ bits1.last)' = bits, false ∈ bits

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits, boolvar, bits2] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: bits, boolvar, bits2