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Proof step 21 from proof bitsand-app-ones (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
bits0 = bits3 + boolvar ', ∀ bits0, bits, bits2. # bits2 < # bits1 → true → bitsand(bits + true ' + bits0, false ' + bits2) = bitsand(true ' + bits0, false ' + bits2) ∨ false ∈ bits ⊦ bits1 = [], bits0 = [], bitsand(bits + true ' + bits0, false ' + bits1) = bitsand(true ' + bits3, false ' + bits1.butlast) + (boolvar ∧ bits1.last)', false ∈ bits

Rule argument

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