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Proof step 38 from proof bitsor-mklist-in (previous step)

bits0 = bits3 + boolvar2 ', ∀ 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 = [], n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits2, bits3) ∨ (boolvar ↔ boolvar1 ∨ boolvar2) ↔ boolvar ∈ bitsor(boolvar0 ' + bits2 + boolvar1 ', bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
∀ bits, boolvar0, boolvar, n, bits1. # bits1 < #(bits3 + boolvar2 ') → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits2, bits3) ∨ (boolvar ↔ boolvar1 ∨ boolvar2) ↔ boolvar ∈ bitsor(boolvar0 ' + bits2 + boolvar1 ', bits3 + boolvar2 '), bits3 = []

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;