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Proof step 2 from proof get-nodups (previous step)

m < # x, n < # x ⊦ x[m] = x[n] ↔ m = n, dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x[m] = x[n], m < # x, n < # x ⊦ m = n, x = [], dups(x)

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic