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Proof step 2 from proof subset-emp (previous step)

H ⊆ ∅ ⊦ H = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ H0 ⊆ H1 ↔ (∀ r. (r ∈ H0 → r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 4:
∀ r. (r ∈ H → r ∈ ∅) ∧ (r ∈ H → H[r] = (∅)[r]) ⊦ H = ∅

Rule argument

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

Additional information

Used terms: ∅, H