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Proof step 6 from proof exnotempty (previous step)

¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0) ⊦ st = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 8:
∀ a, d, st0. ¬(st = st0[a, d] ∧ ¬ a ∈ st0) ⊦ ∀ a. (a ∈ st ↔ a ∈ ∅) ∧ st[a] = (∅)[a]

Rule argument

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

Additional information

Used terms: ∅, st