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Proof step 9 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)][∅ ++ r, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ R[rs, f, g] ↔ R(mk(rs, f, g))
Proof step 11:
r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, r = null, rs = ∅, g(r)[r => f(r)](mk(∅ ++ r, f, g))

Rule argument

spec: heap-fun, inst: heap-aux name: apply seq: ⊦ R[rs, f, g] ↔ R(mk(rs, f, g)) subst: [g, f, rs, R] → [g, f, ∅ ++ r, g(r)[r => f(r)]] flags: <all paths>

Additional information

Used terms: g, f, ∅ ++ r, g(r)[r => f(r)]