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Proof step 19 from proof bitsxor-zeros (previous step)

# bits0 ≤ # bits2 + 1, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ # bits0 = 1, bits0 = [], bitsxor(bits2, bits0.butlast) + (¬(boolvar ↔ bits0.last))' = bits2 + boolvar ', bits2 = [], true ∈ bits0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
bits0 = bits3 + boolvar0 ', # bits0 ≤ # bits2 + 1, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ # bits0 = 1, bits0 = [], bitsxor(bits2, bits3) + (¬(boolvar ↔ boolvar0))' = bits2 + boolvar ', bits2 = [], true ∈ bits0

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits0, boolvar0, 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, boolvar0, bits3