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Proof step 9 from proof pos-at (previous step)

m < # str0, ∀ n. n < # str0 → pos(s→l(str0)[n], str0) = n ⊦ pos(s→l(str0)[m], char ' + str0) = m + 1, char ' + str0 = "", str0 = "", char ∈ s→l(str0), dups(s→l(str0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ char1 ≠ char2 → pos(char1, char2 ' + str) = pos(char1, str) + 1
Proof step 11:
¬ s→l(str0)[m] ≠ char, m < # str0, ∀ n. n < # str0 → pos(s→l(str0)[n], str0) = n ⊦ pos(s→l(str0)[m], char ' + str0) = m + 1, char ' + str0 = "", str0 = "", char ∈ s→l(str0), dups(s→l(str0))
Proof step 14:
s→l(str0)[m] ≠ char, m < # str0, ∀ n. n < # str0 → pos(s→l(str0)[n], str0) = n ⊦ pos(s→l(str0)[m], str0) + 1 = m + 1, char ' + str0 = "", str0 = "", char ∈ s→l(str0), dups(s→l(str0))

Rule argument

spec: string-less inst: name: Pos-rec seq: ⊦ char1 ≠ char2 → pos(char1, char2 ' + str) = pos(char1, str) + 1 subst: [str, char2, char1] → [str0, char, s→l(str0)[m]] flags: rewrite

Additional information

Used terms: str0, char, s→l(str0)[m]