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Proof step 23 from proof sorted-sub-merge (conclusion)

⊦ ⊑(ar[n], ar[n + 1]) ∧ sorted(ar, n + 1, m + 1, ⊑) ∧ sorted(ar, 0, n + 1, ⊑) ∧ transitive(⊑) → (¬ sorted(ar, 0, m + 1, ⊑) ↔ 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-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDFORWARD : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDFORWARD : nat, , fle, ⊦ m < n ∧ n ≤ n0 → m < n0], [CSIMPNAMEDSEQ : nat, , lels, ⊦ ¬ m < n ↔ n ≤ m], [CSIMPNAMEDSEQ : nat, , succ-01, ⊦ m + 1 ≤ n ↔ m < n], [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], [CSIMPNAMEDSEQ : nat, , lez, ⊦ 0 ≤ m])]]