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Proof step 13 from proof owns-one-map (previous step)

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, mk(∅ ++ r, f) = (∅ ++ r)[r, f(r)], r = null, rs = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ((r => o))(H) ↔ r ≠ null ∧ H = (∅ ++ r)[r, o]; (ss)% ⊦ r ∈ ∅ ++ r0 ↔ r = r0; (s)