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Proof step 67 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g(r; a1)], r ∈ rs0, ∀ r0. r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1, ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ o;(∅ ++ r)[r0])) ⊦ # rs0 = 0, r = null, rs0 = ∅, (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], g(r)[r => o][rs1, f, g] ∧ R[rs0, f, g]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: r

Additional information

Used simplifier rules: int-a: ⊦ (rs ∩ rs0) ∩ rs1 = rs ∩ rs0 ∩ rs1; from specification set-union[refset]%int-c: ⊦ rs ∩ rs0 = rs0 ∩ rs; from specification set-union[refset]% ⊦ rs ∩ rs0 = ∅ ∧ r ∈ rs → (r ∈ rs0 ↔ false); (as) Used terms: r