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Proof step 1 from proof prefix-eq (conclusion)

⊦ x ⊑ y → (y ⊑ x ↔ x = y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
y ⊑ x, x ⊑ y ⊦ x = y

Rule argument

Additional information

Used simplifier rules: prefix-trans: ⊦ x ⊑ y ∧ y ⊑ z → x ⊑ z;%prefix-self: ⊦ x ⊑ x;