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Proof step 4 from proof owns-one-map (previous step)

r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, r = null, rs = ∅, (g(r)[r => f(r)] * true)[rs, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ (R0 * R1)[rs, f, g] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ R0[rs0, f, g] ∧ R1[rs1, f, g])
Proof step 6:
r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, r = null, rs = ∅, ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ g(r)[r => f(r)][rs0, f, g] ∧ (true)[rs1, f, g]

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, true, g(r)[r => f(r)]] flags: <all paths>

Additional information

Used terms: g, f, rs, true, g(r)[r => f(r)]