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Proof step 4 from proof disjoint-union-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦ st0 = ∅ ∧ st1 = ∅

Rule argument

Additional information

Used simplifier rules: union-at-left: ⊦ (st0 ∪ st1)[a] = (a ∈ st0 ⊃ st0[a];st1[a]);%union-in: ⊦ a ∈ st0 ∪ st1 ↔ a ∈ st0 ∨ a ∈ st1;% ⊦ ¬ a ∈ ∅; (s)