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Proof step 25 from proof bitsor-mklist-in (previous step)

∀ bits, boolvar0, boolvar, n, bits1. # bits1 < # bits0 → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ # bits0 = 1, bits0 = [], boolvar ∈ bitsor(mklist(boolvar0, m), bits0.butlast) ∨ (boolvar ↔ boolvar0 ∨ bits0.last) ↔ boolvar ∈ bitsor(boolvar0 ', bits0), m = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 26:
bits0 = bits2 + boolvar1 ', ∀ bits, boolvar0, boolvar, n, bits1. # bits1 < # bits0 → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ # bits0 = 1, bits0 = [], boolvar ∈ bitsor(mklist(boolvar0, m), bits2) ∨ (boolvar ↔ boolvar0 ∨ boolvar1) ↔ boolvar ∈ bitsor(boolvar0 ', bits0), m = 0

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits0, boolvar1, 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: bits0, boolvar1, bits2