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Proof step 2 from proof aux-known (previous step)

(a0[r => o] * R)[rs, f, g] ⊦ (a1[r => o] * R)[rs, f, g(r; a1)] ∧ g(r) = a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ (R0 * R1)[rs, f, g] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ R0[rs0, f, g] ∧ R1[rs1, f, g])
Proof step 4:
∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ a0[r => o][rs0, f, g] ∧ R[rs1, f, g] ⊦ (a1[r => o] * R)[rs, f, g(r; a1)] ∧ g(r) = a0

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, rs, R, a0[r => o]] flags: <all paths>

Additional information

Used terms: g, f, rs, R, a0[r => o]