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Proof step 208 from proof insert-correct (previous step)

⟨loop if ¬ found?0 then {if T1[h1(a)] = empty then {T1 := T1[h1(a), full(a)]; found?0 := true}; if ¬ found?0 then {a := T1[h1(a)].elem, T1 := T1[h1(a), T1[h1(a)].elem:= a]; if T2[h2(a)] = empty then {T2 := T2[h2(a), full(a)]; found?0 := true}}; if ¬ found?0 then a := T2[h2(a)].elem, T2 := T2[h2(a), T2[h2(a)].elem:= a]} times kappa⟩ ((T1 = T ∧ (found?0 ↔ found?1) ∧ a = a0 ∧ T2 = T0) ∧ ¬ ¬ found?0), ⟨lookup(a, T1, T2, h1, h2; found?)⟩ (found? ↔ found?0), inv(T1, h1), inv(T2, h2), disj(T1, T2), abs(T1, T2, h1, h2) ++ a ≠ abs(T, T0, h1, h2) ∨ ¬ inv(T, h1) ∨ ¬ inv(T0, h2) ∨ ¬ disj(T, T0) ⊦

Heuristic split applied rule split left

And obtained the following new goals

Rule argument

pos:left 1

Additional information

[Concretesimprules(())]