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Proof step 6 from proof diff-at (previous step)

r ∈ H0, ∀ H, r, H0. # H < # H1 → r ∈ H0 → (H0 \ H)[r] = H0[r] ∨ r ∈ H ⊦ H1 = ∅, (H0 \ H1)[r] = H0[r], r ∈ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 8:
r ∈ H0, ∀ H, r, H0. # H < # H1 → r ∈ H0 → (H0 \ H)[r] = H0[r] ∨ r ∈ H ⊦ (H0 \ H1)[r] = H0[r], r ∈ H1, ∀ r. (r ∈ H1 ↔ r ∈ ∅) ∧ (r ∈ H1 → H1[r] = (∅)[r])

Rule argument

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

Additional information

Used terms: ∅, H1