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Proof step 13 from proof ar-elems-in-h (previous step)

∀ n, ar. n < n1 ∧ n1 ≤ # ar ∧ n < # ar → ar[n] ∈ ar elems n1, n < n1 +1, n1 +1 ≤ # ar, n < # ar ⊦ ar[n] ∈ ar elems n1 +1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : elems-rec, , , n < # ar ⊦ ar elems (n + 1) = (ar elems n) ++ ar[n]], [CSIMPNAMEDSEQ : nat, , lels, ⊦ ¬ m < n ↔ n ≤ m], [CSIMPNAMEDSEQ : nat, , ls-le-simp, ⊦ m < n → (n ≤ m ↔ false)], [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-basic2, , succ, ⊦ m +1 = m + 1])]]