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Proof step 28 from proof and-size-one (previous step)

bits.first, # bits ≤ # bits3 + 1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits = true ', bits = [], bits3 + boolvar ' = (bits3 + boolvar ').first ', bitsand(bits.butlast, bits0 + bits3) = bits0 + bitsand(bits.butlast, bits3), bits.butlast.first

User applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
bits = bits4 + boolvar0 ', bits.first, # bits ≤ # bits3 + 1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits = true ', bits = [], bits3 + boolvar ' = (bits3 + boolvar ').first ', bitsand(bits4, bits0 + bits3) = bits0 + bitsand(bits4, bits3), bits4.first

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits, boolvar0, bits4] 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, boolvar0, bits4