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Proof step 11 from proof bitsor-mklist-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ bits, boolvar0, boolvar, n, bits1. # bits1 < 1 → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits, false ') ↔ boolvar ∈ bitsor(boolvar0 ' + bits, false ')

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ # boolvar ' = 0 +1;% ⊦ # bits = 1 ∧ bits ≠ true ' → bits = false '; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];