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Proof step 21 from proof ex-shift-o (previous step)

P(H0), oP(o)(H1), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ((!(me, O, _∃ oP) * P))(H1 ∪ H0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 23:
P(H0), oP(o)(H1), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ((!(me, O, _∃ oP) * P))(H1 ∪ H0), !(me, O, _∃ oP)(H1) ∧ P(H0) ∧ H1 ⊥ 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, H1, P, !(me, O, _∃ oP)] flags: <all paths>

Additional information

Used terms: H0, H1, P, !(me, O, _∃ oP)