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Proof step 16 from proof bitsand-zeros (previous step)

# bits ≤ 2 + # bits2 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ # bits = 1, bits = [], bitsand(bits.butlast, boolvar ' + bits2) + (bits.last ∧ boolvar0)' = bits, true ∈ bits

User applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
bits = bits3 + boolvar1 ', # bits ≤ 2 + # bits2 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ # bits = 1, bits = [], bitsand(bits3, boolvar ' + bits2) + (boolvar1 ∧ boolvar0)' = bits, true ∈ 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