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Proof step 11 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)](mk(∅ ++ r, f, g))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 13:
r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, r = null, rs = ∅, ((r => f(r)))(mk(∅ ++ r, f)) ∧ (∀ r0. r0 ∈ ∅ ++ r → g(r0) = g(r))

Rule argument

spec: embed-aux name: lift-aux seq: ⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a) subst: [g, f, rs, P, a] → [g, f, ∅ ++ r, r => f(r), g(r)] flags: <all paths>

Additional information

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