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Proof step 5 from proof inj-put (previous step)

∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0[a, d]) ∧ st[a, d][a0] = st0[a, d][a0] ⊦ st = st0, a ∈ st0, a ∈ st

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:
∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0[a, d]) ∧ st[a, d][a0] = st0[a, d][a0] ⊦ ∀ a. (a ∈ st ↔ a ∈ st0) ∧ st[a] = st0[a], a ∈ st0, a ∈ st

Rule argument

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

Additional information

Used terms: st0, st