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Proof step 18 from proof inv-reduce (conclusion)

⊦ inv(T1, h) → (¬ inv(T2, h) ↔ ¬((# T1 = SIZE → # T2 = SIZE) ∧ (∀ n, a. (n < # T1 ∧ T1[n] = full(a) → h(a) = n) → n < # T2 ∧ T2[n] = full(a) → h(a) = n)))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(size,,, ⊦ inv(T, h) → # T = SIZE)))]