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Proof step 14 from proof map-in0 (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: r

Additional information

Used terms: r