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Proof step 3 from proof some-element (previous step)

x = a0 + x0 ⊦ x = [], ∃ a. a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ∃ a. a = a0 ∨ a ∈ x0

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%e-01: ⊦ a ' + x ≠ [];%a: ⊦ (x + y) + z = x + y + z;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);