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Proof step 15 from proof diffgetn-nodupssize (previous step)

n = m0 + 1, n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ 1 + #(x0 \ x0[m0] ') = # x0, a = x0[m0], n = 0, x0 = [], dups(x0), a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m0 < # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ 1 + #(x0 \ x0[m0] ') = # x0, a = x0[m0], x0 = [], dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)