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Proof step 25 from proof subarray-append-shift-right (previous step)

m ≤ # ar0 + # ar1, # ar0 ≤ m ⊦ subarray(ar0 + ar1, 0, m) = ar0 + subarray(ar1, 0, m - # ar0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
m = n0 + # ar0, m ≤ # ar0 + # ar1, # ar0 ≤ m ⊦ subarray(ar0 + ar1, 0, m) = ar0 + subarray(ar1, 0, n0)

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, # ar0, n] flags:

Additional information

[Concretesimprules((Csimpelim(n ≤ m ⊦ n0 = m - n ↔ m = n0 + n))), Extraterms((m# ar0n))]