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Proof step 1 from proof ni (conclusion)

⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + a ' = y + z, a ∈ y ⊦ z = [], a ∈ z

Rule argument

Additional information

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