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Proof step 11 from proof ar-elems-append-not-lower-h (conclusion)

⊦ n > # ar0 ∧ n ≤ # ar0 + # ar1 → (ar0 + ar1) elems n = (ar0 elems # ar0) ∪ (ar1 elems (n - # ar0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# ar0 < n, n ≤ # ar0 + # ar1 ⊦ (ar0 + ar1) elems n = (ar0 elems # ar0) ∪ (ar1 elems (n - # ar0)), # ar1 = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : nat-basic2, , c, ⊦ m + n = n + m], [CSIMPNAMEDSEQ : nat-basic2, , a, ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPNAMEDSEQ : nat-basic2, , l-01, ⊦ m < m + n ↔ n ≠ 0], [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, , ax-02, ⊦ m > n ↔ n < m])]]