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Proof step 6 from proof pos-at (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # str0, ∀ n0. n0 < # str0 → pos(s→l(str0)[n0], str0) = n0 ∨ str0 = "" ⊦ char ' + str0 = "", n = 0, pos(s→l(str0)[n - 1], char ' + str0) = n, dups(s→l(str0)), char ∈ s→l(str0)

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1;