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Proof step 51 from proof bitsxor-is-orand (previous step)

∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ # bits = 1, bits2 = [], bits = [], bits2int(bitsxor(bits, bits2 + boolvar ')) = bits2int (bitsor (bitsand(bitsinvert(bits).butlast, bits2) + (bitsinvert(bits).last ∧ boolvar)', bitsand(bitsinvert(bits2), bits.butlast) + (¬ boolvar ∧ bits.last)'))

User applied rule insert elim lemma

And obtained the following new goals

Proof step 52:
bits = bits0 + boolvar0 ', ∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ # bits = 1, bits2 = [], bits = [], bits2int(bitsxor(bits, bits2 + boolvar ')) = bits2int (bitsor (bitsand(bitsinvert(bits).butlast, bits2) + (bitsinvert(bits).last ∧ boolvar)', bitsand(bitsinvert(bits2), bits0) + (¬ boolvar ∧ boolvar0)'))

Rule argument

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