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Proof step 29 from proof insert-correct (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 12:
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) ⊦

Rule argument

pos:left 6

Additional information

[Concretesimprules(())]