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Proof step 16 from proof owns-one-map (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]); (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)% ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ H[r0, o] ↔ r ∈ H; (s)% ⊦ r ∈ ∅ ++ r0 ↔ r = r0; (s)% ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; (s)