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Proof step 2 from proof inapp-02 (previous step)

⊦ y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
y + c ' + z = a ' + z0, a ∈ c ' ∨ a ∈ z ⊦ a ∈ y, y = [] ∧ c = a ∧ z = z0

Rule argument

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%app-07: ⊦ a ' + x = b ' + x ↔ a = b;%in-14: ⊦ a ' + x = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;