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Proof step 11 from proof k-Extension (conclusion)

⊦ st0 ≠ st1 ↔ ¬(∀ a. (a ∈ st0 ↔ a ∈ st1) ∧ (a ∈ st0 → st0[a] = st1[a]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ a. (a ∈ st0 ↔ a ∈ st1) ∧ (a ∈ st0 → st0[a] = st1[a]) ⊦ st0 = st1

Rule argument

Additional information

[Concretesimprules(())]