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Proof step 2 from proof in-append0-not-left (previous step)

a ' + x = x0 + x1 ⊦ a ∈ x0 ↔ x0 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ' + x = x0 + x1, a ∈ x1 ⊦ x0 = [], a ∈ x0

Rule argument

Additional information

Used simplifier rules: in-empty: ⊦ ¬ a ∈ [];%in-14: ⊦ a ' + x = y → a ∈ y;%a: ⊦ (x + y) + z = x + y + z;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;