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Proof step 4 from proof pos-at (previous step)

n ≤ # str0, ∀ n0. n0 < # str0 → pos(s→l(str0)[n0], str0) = n0 ∨ str0 = "" ⊦ pos((char ' + s→l(str0))[n], char ' + str0) = n, char ' + str0 = "", dups(s→l(str0)), char ∈ s→l(str0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
n ≠ 0 ⊦ (char ' + chars)[(* n)] = chars[n -1]
Proof step 6:
n ≠ 0, n ≤ # str0, ∀ n0. n0 < # str0 → pos(s→l(str0)[n0], str0) = n0 ∨ str0 = "" ⊦ pos(s→l(str0)[n -1], char ' + str0) = n, char ' + str0 = "", dups(s→l(str0)), char ∈ s→l(str0)

Rule argument

spec: list-del inst: charlist name: get-numrec seq: n ≠ 0 ⊦ (char ' + chars)[(* n)] = chars[n -1] subst: [n, chars, char] → [n, s→l(str0), char] flags: rewrite

Additional information

Used simplifier rules: ⊦ (char ' + chars)[0] = char;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ pos(char, char ' + str) = 0;% ⊦ (str1 + str2) + str3 = str1 + str2 + str3; Used terms: n, s→l(str0), char