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Proof step 2 from proof inapp-14 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: app: ⊦ x + [] = x;%a: ⊦ (x + y) + z = x + y + z;%app-03: ⊦ x + y = z + y ↔ x = z;%in-15: ⊦ x + a ' = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;