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Proof step 12 from proof insert-correct (previous step)

abs(T1, T0, h1, h2) ++ a ≠ abs(T1[h1(a), full(a)], T0, h1, h2), T1[h1(a)] = empty, T0[h2(a)] ≠ full(a), inv(T1, h1), inv(T0, h2), disj(T1, T0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
T1[h1(a)] = empty, disj(T1, T0), inv(T0, h2), inv(T1, h1) ⊦ abs(T1, T0, h1, h2) ++ a = abs(T1[h1(a), full(a)], T0, h1, h2), T0[h2(a)] = full(a)

Rule argument

Additional information

[Concretesimprules(())]