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Proof step 6 from proof merge-owned (previous step)

rs1 ∩ rs0 = ∅, a[P][rs1, f, g], R[rs0, f, g], (a0[P0] * R0)[rs1 ∪ rs0, f, g] ⊦ a0 = a, (a[P] * a0[P0] * true)[rs1 ∪ rs0, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ (R0 * R1)[rs, f, g] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ R0[rs0, f, g] ∧ R1[rs1, f, g])
Proof step 8:
rs1 ∩ rs0 = ∅, R[rs0, f, g], a[P][rs1, f, g], ∃ rs3, rs2. rs3 ∪ rs2 = rs1 ∪ rs0 ∧ rs3 ∩ rs2 = ∅ ∧ a0[P0][rs3, f, g] ∧ R0[rs2, f, g] ⊦ a0 = a, (a[P] * a0[P0] * true)[rs1 ∪ rs0, f, g]

Rule argument

spec: heap-fun, inst: heap-aux name: sep seq: ⊦ (R0 * R1)[rs, f, g] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ R0[rs0, f, g] ∧ R1[rs1, f, g]) subst: [g, f, rs, R1, R0] → [g, f, rs1 ∪ rs0, R0, a0[P0]] flags: <all paths>

Additional information

Used terms: g, f, rs1 ∪ rs0, R0, a0[P0]