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Proof step 5 from proof disjoint-union-empty (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 7:
∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦ (∀ a. (a ∈ st0 ↔ a ∈ ∅) ∧ st0[a] = (∅)[a]) ∧ st1 = ∅

Rule argument

spec: store name: Extension seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]) subst: [st2, st1] → [∅, st0] flags: <all paths>

Additional information

Used terms: ∅, st0