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Proof step 1 from proof at-diff (conclusion)

⊦ @(n, m0, m1) ∈ st ∧ ¬ n ∈ ns → (st \ ns)[@(n, m0, m1)] = st[@(n, m0, m1)]

System applied rule normalize

And obtained the following new goals

Proof step 2:
@(n, m0, m1) ∈ st ⊦ (st \ ns)[@(n, m0, m1)] = st[@(n, m0, m1)], n ∈ ns

Rule argument

Additional information