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Proof step 2 from proof substring-app-rec (previous step)

# str ≤ m ⊦ substring(str + str0, m, n) = substring(str0, m - # str, n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
m = n0 + # str, # str ≤ m ⊦ substring(str + str0, m, n) = substring(str0, n0, n)

Rule argument

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

Additional information

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