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Proof step 6 from proof mod-other-both (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ R0[rs0, f, g] ∧ R1[rs1, f, g] ∧ rs0 ∩ rs1 = ∅ → (R0 * R1)[rs0 ∪ rs1, f, g]
Proof step 8:
rs1 ∩ rs0 = ∅, R[rs0, f, g], a[P][rs1, f, g] ⊦ g(r) = a, (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)], a[P][rs1, f(r; o), g(r; a0)] ∧ (true)[rs0, f(r; o), g(r; a0)] ∧ rs1 ∩ rs0 = ∅

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; a0), f(r; o), rs0, rs1, true, a[P]] flags: <all paths>

Additional information

Used terms: g(r; a0), f(r; o), rs0, rs1, true, a[P]