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Proof step 5 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, a0[r => o][rs1, f, g], R[rs0, f, g] ⊦ (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ R0[rs0, f, g] ∧ R1[rs1, f, g] ∧ rs0 ∩ rs1 = ∅ → (R0 * R1)[rs0 ∪ rs1, f, g]
Proof step 7:
rs1 ∩ rs0 = ∅, R[rs0, f, g], a0[r => o][rs1, f, g] ⊦ (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, a1[r => o][rs1, f, g(r; a1)] ∧ R[rs0, f, g(r; a1)] ∧ rs1 ∩ rs0 = ∅
Proof step 31:
rs1 ∩ rs0 = ∅, R[rs0, f, g], a0[r => o][rs1, f, g], a1[r => o][rs1, f, g(r; a1)], R[rs0, f, g(r; a1)] ⊦ true ∧ g(r) = a0

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(r; a1), f, rs0, rs1, R, a1[r => o]] flags: <all paths>

Additional information

Used terms: g(r; a1), f, rs0, rs1, R, a1[r => o]