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Proof step 24 from proof and-onesin (previous step)

∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits0 = [], bits = [], (boolvar ↔ bits.last ∧ bits0.last) ∨ boolvar ∈ bitsand(bits.butlast, bits0.butlast) ↔ boolvar ∈ bits0, bits = bits.first ', bits0 = bits0.first ', false ∈ bits

User applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
bits0 = bits2 + boolvar0 ', ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits0 = [], bits = [], (boolvar ↔ bits.last ∧ boolvar0) ∨ boolvar ∈ bitsand(bits.butlast, bits2) ↔ boolvar ∈ bits0, bits = bits.first ', bits0 = bits0.first ', false ∈ bits

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits0, boolvar0, 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: bits0, boolvar0, bits2