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Proof step 2 from proof inapp-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

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;%isin: ⊦ a ' + x = y + b ' + z ∧ y ≠ [] → b ∈ x;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;