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Proof step 26 from proof and-onesin (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 27:
bits = bits3 + boolvar1 ', ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits = bits.first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits = [], (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits3, bits2) ↔ (boolvar ↔ boolvar0) ∨ boolvar ∈ bits2, false ∈ bits

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits, boolvar1, bits3] 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, boolvar1, bits3