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Proof step 3 from proof subset-at (previous step)

r ∈ H0, H0 ⊆ H1 ⊦ H1[r] = H0[r]

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:
r ∈ H0, ∀ r. (r ∈ H0 → r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]) ⊦ H1[r] = H0[r]

Rule argument

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

Additional information

Used terms: H1, H0