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Proof step 59 from proof splice-resize-less (previous step)

# ar1 < n1 + n2, n < n2 + n1, max(# ar1, n1 + m0) < n2 + n1, n < n1 + n2, n1 ≤ n, max(# ar1, n1 + m0) ≤ n, n2 ≤ m0 ⊦ m0 = 0, n2 = 0, zero = ar0[(n - n1) + n0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 60:
n = n3 + n1, n < n1 + n2, max(# ar1, n1 + m0) < n2 + n1, n < n2 + n1, # ar1 < n1 + n2, n2 ≤ m0, max(# ar1, n1 + m0) ≤ n, n1 ≤ n ⊦ m0 = 0, n2 = 0, zero = ar0[n3 + n0]

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n1, n3]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n1, n3