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Proof step 2 from proof disjoint-union-empty (previous step)

st0 ∪ st1 = ∅ ⊦ st0 = ∅ ∧ st1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 4:
∀ a. (a ∈ st0 ∪ st1 ↔ a ∈ ∅) ∧ (st0 ∪ st1)[a] = (∅)[a] ⊦ st0 = ∅ ∧ st1 = ∅

Rule argument

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

Additional information

Used terms: ∅, st0 ∪ st1