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Proof step 13 from proof diffgetn-nodupssize (previous step)

n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ 1 + #(x0 \ x0[n - # a '] ') = # x0, a = x0[n - # a '], x0 = [], n < # a ', dups(x0), a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ 1 + #(x0 \ x0[n - 1] ') = # x0, a = x0[n - 1], n = 0, x0 = [], dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ m < 1 ↔ m = 0; (s)% ⊦ # a ' = 0 +1; (s)