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Proof step 4 from proof sorted-sub-entire (conclusion)

n = # ar ⊦ sorted(ar, 0, n, ⊑) → (¬ sorted(ar, ⊑) ↔ false)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : sorted-entire, , , ⊦ sorted(ar, ⊑) ↔ (∀ n, m. n < m ∧ m < # ar → ⊑(ar[n], ar[m]))], [CSIMPNAMEDSEQ : sorted-sub, , , ⊦ sorted(ar, n0, n1, ⊑) ↔ (∀ n, m. n0 ≤ n ∧ n < m ∧ m < n1 → ⊑(ar[n], ar[m]))], [CSIMPNAMEDSEQ : nat-basic2, , succ, ⊦ m +1 = m + 1], [CSIMPNAMEDSEQ : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDFORWARD : nat-basic1, , trans-1, ⊦ n < n0 ∧ n0 < n1 → n < n1], [CSIMPNAMEDFORWARD : nat-basic1, , lf, ⊦ m < n ∧ n < n0 → m +1 < n0], [CSIMPNAMEDFORWARD : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDSEQ : nat, , lez, ⊦ 0 ≤ m])]]