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Proof step 3 from proof sorted-sub-less1 (conclusion)

⊦ sorted(ar, m, n + 1, ⊑) → (¬ sorted(ar, m, n, ⊑) ↔ false)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([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, , fle-01, ⊦ m ≤ n ∧ n < n0 → m < n0], [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-basic2, , c, ⊦ m + n = n + m], [CSIMPNAMEDSEQ : nat-basic2, , a, ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPNAMEDSEQ : nat-basic2, , s, ⊦ m < n + 1 ↔ ¬ n < m])]]