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Proof step 5 from proof exin (previous step)

r ∈ H ⊦ H = ((H -- r) ++ r)[r, H[r]]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 7:
r ∈ H ⊦ ∀ r0. (r0 ∈ H ↔ r0 ∈ ((H -- r) ++ r)[r, H[r]]) ∧ (r0 ∈ H → H[r0] = ((H -- r) ++ r)[r, H[r]][r0])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [((H -- r) ++ r)[r, H[r]], H] flags: <all paths>

Additional information

Used terms: ((H -- r) ++ r)[r, H[r]], H