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Proof step 2 from proof sorted-sub-change-above (conclusion)

n ≥ m1 ⊦ sorted(ar[n, a], m0, m1, ⊑) ↔ sorted(ar, m0, m1, ⊑)

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 : array, , cngetput, m ≠ n ⊦ (ar[m, a][n]) = ar[n]], [CSIMPNAMEDSEQ : nat, , ge, ⊦ n ≥ m ↔ m ≤ n], [CSIMPNAMEDFORWARD : nat, , fle, ⊦ m < n ∧ n ≤ n0 → m < n0], [CSIMPNAMEDFORWARD : nat-basic1, , lf, ⊦ m < n ∧ n < n0 → m +1 < n0], [CSIMPNAMEDFORWARD : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDFORWARD : nat-basic1, , trans-1, ⊦ n < n0 ∧ n0 < n1 → n < n1], [CSIMPNAMEDFORWARD : nat, , fle-01, ⊦ m ≤ n ∧ n < n0 → m < n0], [CSIMPNAMEDSEQ : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDSEQ : nat-basic1, , ls-02, ⊦ m < n → ¬ n < m])]]