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Proof step 28 from proof bitsor-mklist-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
boolvar ∈ bitsor(mklist(boolvar0, m), bits2) ↔ boolvar ∈ bitsor(boolvar0 ', bits2), ∀ bits, boolvar0, boolvar, n, bits1. # bits1 < #(bits2 + boolvar1 ') → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ boolvar ∈ bitsor(boolvar0 ', bits2) ∨ (boolvar ↔ boolvar0 ∨ boolvar1) ↔ boolvar ∈ bitsor(boolvar0 ', bits2 + boolvar1 '), m = 0, bits2 = []

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;