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Proof step 4 from proof exa (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a, x0

Additional information

Used terms: a, x0