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Proof step 5 from proof owns-none-one2 (previous step)

∀ r. r ∈ rs → g(r) ≠ a ⊦ r = null, owns-one(a, r, rs ++ r, g(r; a)), r ∈ rs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ owns-one(a, r, rs, g) ↔ r ≠ null ∧ r ∈ rs ∧ g(r) = a ∧ (∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ a)
Proof step 7:
∀ r. r ∈ rs → g(r) ≠ a ⊦ r = null, r ∈ rs, r ≠ null ∧ r ∈ rs ++ r ∧ g(r; a)(r) = a ∧ (∀ r0. r0 ∈ rs ++ r ∧ r0 ≠ r → g(r; a)(r0) ≠ a)

Rule argument

spec: <toplevel> name: owns-one seq: ⊦ owns-one(a, r, rs, g) ↔ r ≠ null ∧ r ∈ rs ∧ g(r) = a ∧ (∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ a) subst: [g, rs, r, a] → [g(r; a), rs ++ r, r, a] flags: <all paths>

Additional information

Used terms: g(r; a), rs ++ r, r, a