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Proof step 2 from proof sorted-1 (conclusion)

⊦ sorted(ar, n, n + 1, ⊑)

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, , ls-le-simp, ⊦ m < n → (n ≤ m ↔ false)], [CSIMPNAMEDSEQ : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDFORWARD : nat, , fle-01, ⊦ m ≤ n ∧ n < n0 → m < n0], [CSIMPNAMEDFORWARD : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDFORWARD : nat, , fle, ⊦ m < n ∧ n ≤ n0 → m < n0], [CSIMPNAMEDFORWARD : nat, , let, ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1], [CSIMPNAMEDSEQ : nat, , lels, ⊦ ¬ m < n ↔ n ≤ m], [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])]]