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Proof step 2 from proof ext-rec (previous step)

a ∈ st1 ∧ st1[a] = d ∧ (∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b])) ⊦ st1 = st0[a, d]

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:
a ∈ st1 ∧ st1[a] = d ∧ (∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b])) ⊦ ∀ a0. (a0 ∈ st1 ↔ a0 ∈ st0[a, d]) ∧ st1[a0] = st0[a, d][a0]

Rule argument

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

Additional information

Used terms: st0[a, d], st1