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Proof step 10 from proof and-size-one-01 (previous step)

# bits3 + 1 = # bits1, ∀ bits, bits1, bits0. # bits < #(bits3 + boolvar ') → # bits = # bits1 → bits1 = [] ∨ bits0 = [] ∨ bitsand(true ' + bits, bits0 + bits1) = bits0 + bitsand(bits, bits1) ⊦ bits1 = [], bits0 = [], bitsand(true ' + bits3, bits0 + bits1.butlast) + (boolvar ∧ bits1.last)' = bits0 + bitsand(bits3 + boolvar ', bits1), # bits0 + # bits1 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
bits1 = bits2 + boolvar0 ', # bits3 + 1 = # bits1, ∀ bits, bits1, bits0. # bits < #(bits3 + boolvar ') → # bits = # bits1 → bits1 = [] ∨ bits0 = [] ∨ bitsand(true ' + bits, bits0 + bits1) = bits0 + bitsand(bits, bits1) ⊦ bits1 = [], bits0 = [], bitsand(true ' + bits3, bits0 + bits2) + (boolvar ∧ boolvar0)' = bits0 + bitsand(bits3 + boolvar ', bits1), # bits0 + # bits1 = 0

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits1, 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: bits1, boolvar0, bits2