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Proof step 14 from proof merge-owned (previous step)

rs1 ∩ rs3 = ∅, rs1 ∩ rs2 = ∅, rs2 ∩ rs3 = ∅, rs2 ∩ rs = ∅, a0[P0][rs2, f, g], R0[rs, f, g], a[P][rs1, f, g], R[rs2 ∪ rs3, f, g] ⊦ a0 = a, (a[P] * a0[P0] * true)[rs1 ∪ rs2 ∪ rs3, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ R0[rs0, f, g] ∧ R1[rs1, f, g] ∧ rs0 ∩ rs1 = ∅ → (R0 * R1)[rs0 ∪ rs1, f, g]
Proof step 16:
rs2 ∩ rs = ∅, rs2 ∩ rs3 = ∅, rs1 ∩ rs2 = ∅, rs1 ∩ rs3 = ∅, R[rs2 ∪ rs3, f, g], a[P][rs1, f, g], R0[rs, f, g], a0[P0][rs2, f, g] ⊦ a0 = a, (a[P] * a0[P0] * true)[rs1 ∪ rs2 ∪ rs3, f, g], a[P][rs1, f, g] ∧ (a0[P0] * true)[rs2 ∪ rs3, f, g] ∧ rs1 ∩ (rs2 ∪ rs3) = ∅

Rule argument

spec: heap-fun, inst: heap-aux name: split-uni seq: ⊦ R0[rs0, f, g] ∧ R1[rs1, f, g] ∧ rs0 ∩ rs1 = ∅ → (R0 * R1)[rs0 ∪ rs1, f, g] subst: [g, f, rs1, rs0, R1, R0] → [g, f, rs2 ∪ rs3, rs1, a0[P0] * true, a[P]] flags: <all paths>

Additional information

Used terms: g, f, rs2 ∪ rs3, rs1, a0[P0] * true, a[P]