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Proof step 1 from proof oplus-shift-put-l (conclusion)

⊦ st0[a, d] ⊕ st1 = (st0 ⊕ st1)[a, d]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ (a ∈ st1 → st1[a] = st2[a]))
Proof step 3:
⊦ ∀ a0. (a0 ∈ st0[a, d] ⊕ st1 ↔ a0 ∈ (st0 ⊕ st1)[a, d]) ∧ (a0 ∈ st0[a, d] ⊕ st1 → (st0[a, d] ⊕ st1)[a0] = (st0 ⊕ st1)[a, d][a0])

Rule argument

spec: store inst: name: k-Extension seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ (a ∈ st1 → st1[a] = st2[a])) subst: [st2, st1] → [(st0 ⊕ st1)[a, d], st0[a, d] ⊕ st1] flags: rewrite

Additional information

Used terms: (st0 ⊕ st1)[a, d], st0[a, d] ⊕ st1