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Proof step 3 from proof exa (previous step)

x0 = [] ∨ (∃ a, y. x0 = a + y) ⊦ a + x0 = [], ∃ a0, y. a + x0 = a0 + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x0 = [] ∨ x0 = a0 + y ⊦ ∃ a0, y. a + x0 = a0 + y

Rule argument

Additional information

Used simplifier rules: disj-1: ⊦ [] ≠ a + x;