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Proof step 12 from proof ex-shift-o-p (previous step)

P(H0), P0(H), oP(o)(H2), H0 ⊥ H, H0 ⊥ H2, H2 ⊥ H, ∀ r. r ∈ H2 ∨ r ∈ H → O(r) = me ⊦ ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0), ∃ o. ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 14:
P(H0), P0(H), oP(o)(H2), H2 ⊥ H, H0 ⊥ H2, H0 ⊥ H, ∀ r. r ∈ H2 ∨ r ∈ H → O(r) = me ⊦ ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0), ∃ o. ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0), !(me, O, oP(o) * P0)(H2 ∪ H) ∧ P(H0) ∧ H2 ∪ H ⊥ H0

Rule argument

spec: heap-sep name: split-union seq: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1) subst: [H1, H0, P1, P0] → [H0, H2 ∪ H, P, !(me, O, oP(o) * P0)] flags: <all paths>

Additional information

Used terms: H0, H2 ∪ H, P, !(me, O, oP(o) * P0)