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Proof step 10 from proof disjoint-union-empty (previous step)

∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦ ¬ a ∈ st0 ∧ (∀ a. (a ∈ st1 ↔ a ∈ ∅) ∧ st1[a] = (∅)[a])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦ ¬ a ∈ st0 ∧ ¬ a0 ∈ st1

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ st → st[a] = (∅)[a]; (s)% ⊦ ¬ a ∈ ∅; (s)