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Proof step 19 from proof get-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1, ∀ r. r ∈ rs1 → g(r) = a, ∀ r0. (r0 ∈ rs1 → r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ f(r);(∅ ++ r)[r0])) ⊦ # rs1 = 0, g(r) = a, r = null, rs1 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: r

Additional information

Used terms: r