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Proof step 9 from proof ar-elems-append-not-lower-h (previous step)

n = n0 + # ar0 ∧ # ar0 ≤ n, # ar0 < n, n ≤ # ar0 + # ar1 ⊦ (ar0 + ar1) elems n = (ar0 elems # ar0) ∪ (ar1 elems n0), # ar1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 ≤ # ar1 ⊦ (ar0 + ar1) elems (n0 + # ar0) = (ar0 elems # ar0) ∪ (ar1 elems n0), # ar1 = 0, n0 = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : nat, , lep-02, ⊦ m + n ≤ m + n0 ↔ n ≤ n0], [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, , lels-03, ⊦ m < n → m ≤ n])]]