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Proof step 15 from proof bitsor-first (previous step)

bitsor(true ' + bits.butlast, bits0.butlast) + (bits.last ∨ bits0.last)' = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ # bits0 = 1, bits = [], bits0 = [], (bits.last ∨ bits0.last) ∧ ¬ true ∈ bits1, ¬ false ∈ bitsor(true ' + bits.butlast, bits0.butlast) ∧ (bits.last ∨ bits0.last)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
bits = bits3 + boolvar ', bitsor(true ' + bits3, bits0.butlast) + (boolvar ∨ bits0.last)' = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ # bits0 = 1, bits = [], bits0 = [], (boolvar ∨ bits0.last) ∧ ¬ true ∈ bits1, ¬ false ∈ bitsor(true ' + bits3, bits0.butlast) ∧ (boolvar ∨ bits0.last)

Rule argument

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