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Proof step 3 from proof map-putex (previous step)

((r => o1 * P))(H[r, o1]), r ∈ H ⊦ ∃ o. ((r => o * P))(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ r ≠ null → (r ∈ H ↔ (∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0))
Proof step 5:
((r => o1 * P))(H[r, o1]), ∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0 ⊦ r = null, ∃ o. ((r => o * P))(H)

Rule argument

spec: heap 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: ⊦ ¬ null ∈ H; (s) Used terms: H, r