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Proof step 5 from proof exin-uni (previous step)

r ∈ H ⊦ ∃ H0, o. H = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ r ≠ null → (r ∈ H ↔ (∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0))
Proof step 7:
∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0 ⊦ r = null, ∃ H0, o. H = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0

Rule argument

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

Additional information

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