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Proof step 14 from proof ex-lst-shift (previous step)

rs1 ∩ rs0 = ∅, R[rs0, cf, af], xv(x)(mk(rs1, cf)), ∀ r. r ∈ rs1 → af(r) = o ⊦ (o[xv(x)] * R)[rs1 ∪ rs0, cf, af]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ R0[rs0, cf, af] ∧ R1[rs1, cf, af] ∧ rs0 ∩ rs1 = ∅ → (R0 * R1)[rs0 ∪ rs1, cf, af]
Proof step 16:
rs1 ∩ rs0 = ∅, R[rs0, cf, af], xv(x)(mk(rs1, cf)), ∀ r. r ∈ rs1 → af(r) = o ⊦ (o[xv(x)] * R)[rs1 ∪ rs0, cf, af], o[xv(x)][rs1, cf, af] ∧ R[rs0, cf, af] ∧ rs1 ∩ rs0 = ∅

Rule argument

spec: heap-fun, inst: cell-heap-aux-heap-aux name: split-uni seq: ⊦ R0[rs0, cf, af] ∧ R1[rs1, cf, af] ∧ rs0 ∩ rs1 = ∅ → (R0 * R1)[rs0 ∪ rs1, cf, af] subst: [af, cf, rs1, rs0, R1, R0] → [af, cf, rs0, rs1, R, o[xv(x)]] flags: <all paths>

Additional information

Used terms: af, cf, rs0, rs1, R, o[xv(x)]