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Proof step 3 from proof inj-put-01 (previous step)

⊦ st[a, d][b, d0] = st[b, d0] ↔ a ∈ st ∧ st[a] = d, a = b

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 5:
⊦ (∀ a0. (a0 ∈ st[a, d][b, d0] ↔ a0 ∈ st[b, d0]) ∧ st[a, d][b, d0][a0] = st[b, d0][a0]) ↔ a ∈ st ∧ st[a] = d, a = b

Rule argument

spec: <toplevel> name: Extension seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]) subst: [st2, st1] → [st[b, d0], st[a, d][b, d0]] flags: <all paths>

Additional information

Used terms: st[b, d0], st[a, d][b, d0]