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Proof step 34 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)) ⊦ bits = [], bits0 = [], n = 0, boolvar ∈ bitsor((mklist(boolvar0, n) + bits).butlast, bits0.butlast) ∨ (boolvar ↔ (mklist(boolvar0, n) + bits).last ∨ bits0.last) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits0), n + # bits = 0, # bits0 = 1, bits = [] ∧ n = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
∀ bits, boolvar0, boolvar, n, bits1. # bits1 < # bits0 → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ n + # bits = 0, # bits0 = 1, bits = [], bits0 = [], n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits.butlast, bits0.butlast) ∨ (boolvar ↔ bits.last ∨ bits0.last) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits0 + bits).last ↔ bits.last);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;