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Proof step 13 from proof map-in0 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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