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Proof step 3 from proof union-move-left (previous step)

a ∈ st1 ⊦ st0[a, st1[a]] ∪ (st1 -- a) = st0 ∪ st1, a ∈ st0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 5:
a ∈ st1 ⊦ a ∈ st0, ∀ a0. (a0 ∈ st0[a, st1[a]] ∪ (st1 -- a) ↔ a0 ∈ st0 ∪ st1) ∧ (st0[a, st1[a]] ∪ (st1 -- a))[a0] = (st0 ∪ st1)[a0]

Rule argument

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

Additional information

Used terms: st0 ∪ st1, st0[a, st1[a]] ∪ (st1 -- a)