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Proof step 2 from proof pos-get-01 (previous step)

x[n] = a, n < # x ⊦ pos(a, x) = n, dups(x)

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ dups(x) ∧ n < # x → pos(x[n], x) = n

Rule argument

spec: <toplevel> name: pos-get seq: ⊦ ¬ dups(x) ∧ n < # x → pos(x[n], x) = n subst: [x, n] → [x, n]

Additional information

Used simplifier rules: pos-get: ⊦ ¬ dups(x) ∧ n < # x → pos(x[n], x) = n; Used terms: x, n