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Proof step 6 from proof exnotempty (previous step)

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

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

Rule argument

spec: <toplevel> name: Extension seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]) subst: [st2, st1] → [∅, st] flags: <all paths>

Additional information

Used terms: ∅, st