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Proof step 8 from proof diff-in (previous step)

r0 ∈ H1, ∀ H, r, H0. # H < # H1 → true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ r ≠ null → (r ∈ H ↔ (∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0))
Proof step 10:
∃ H0, o. H1 = (H0 ++ r0)[r0, o] ∧ ¬ r0 ∈ H0, ∀ H, r, H0. # H < # H1 → true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1, r0 = null

Rule argument

spec: <toplevel> name: exin seq: ⊦ r ≠ null → (r ∈ H ↔ (∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0)) subst: [H, r] → [H1, r0] flags: <all paths>

Additional information

Used simplifier rules: sanity: ⊦ ¬ null ∈ H; Used terms: H1, r0