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Proof step 22 from proof subarray-split-at (previous step)

n < # ar ⊦ ar = subarray(ar, 0, n) + subarray(ar, n, # ar - n), # ar = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# ar, n, n0] flags:

Additional information

[Concretesimprules((Csimpnamedseq(append-prefix-left,,, ⊦ ar0 + ar1 = ar0 ↔ # ar1 = 0) Csimpnamedseq(append-a,,, ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1) Csimpnamedseq(nat,,lels-01, ⊦ ¬ m ≤ n ↔ n < m) Csimpnamedseq(nat-basic1,,ls-02, ⊦ m < n → ¬ n < m) Csimpnamedseq(array-copy,,subarray-elim, # ar = n ⊦ subarray(ar, 0, n) = ar) Csimpnamedseq(array-copy,,size-subarray, ⊦ # subarray(ar, m, n) = n) Csimpnamedseq(nat,,zero-01, ⊦ n < m → m - n ≠ 0) Csimpelim(n ≤ m ⊦ n0 = m - n ↔ m = n0 + n))), Extraterms((# arnn0))]