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Proof step 20 from proof ex-shift-o-p (previous step)

((!(me, O, oP(o) * P0) * P))(H) ⊦ ((!(me, O, _∃ oP * P0) * P))(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 22:
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ !(me, O, oP(o) * P0)(H0) ∧ P(H1) ⊦ ((!(me, O, _∃ oP * P0) * P))(H)

Rule argument

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

Additional information

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