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Proof step 2 from proof delete-put-rew (previous step)

⊦ (st -- b)[a, d] = st[a, d] -- b, a = b

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:
⊦ ∀ a0. (a0 ∈ (st -- b)[a, d] ↔ a0 ∈ st[a, d] -- b) ∧ (st -- b)[a, d][a0] = (st[a, d] -- b)[a0], a = b

Rule argument

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

Additional information

Used terms: st[a, d] -- b, (st -- b)[a, d]