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Proof step 45 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g(r; a1)], a1[r => f(r)][rs1, f, g(r; a1)], R[rs0, f, g], ∀ r0. (r0 ∈ rs1 → r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ f(r);(∅ ++ r)[r0])), ∀ r. r ∈ rs1 → g(r) = a0 ⊦ g(r) = a0, r = null

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: r

Additional information

Used simplifier rules: map-in0: ⊦ a[r => o][rs, f, g] → r ∈ rs; Used terms: r