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Proof step 3 from proof sw-put (previous step)

r0 ∈ H, r1 ∈ H ⊦ H[r0, o0][r1, o1] = H[r1, o1][r0, o0], r0 = r1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 5:
r1 ∈ H, r0 ∈ H ⊦ r0 = r1, ∀ r. (r ∈ H[r0, o0][r1, o1] ↔ r ∈ H[r1, o1][r0, o0]) ∧ (r ∈ H[r0, o0][r1, o1] → H[r0, o0][r1, o1][r] = H[r1, o1][r0, o0][r])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [H[r1, o1][r0, o0], H[r0, o0][r1, o1]] flags: <all paths>

Additional information

Used terms: H[r1, o1][r0, o0], H[r0, o0][r1, o1]