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Proof step 3 from proof diffgetn-nodupssize (previous step)

n < # x ⊦ #(x \ x[n] ') = # x - 1, x = [], dups(x)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# x = m + 1, n < # x ⊦ #(x \ x[n] ') = m, # x = 0, x = [], dups(x)

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# x, m]

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # x, m