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Proof step 3 from proof sep (previous step)

((P0 * P1))(mk(rs, f)) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 5:
∃ H0, H1. H0 ⊥ H1 ∧ mk(rs, f) = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

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] → [mk(rs, f), P1, P0] flags: <all paths>

Additional information

Used terms: mk(rs, f), P1, P0