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Proof step 9 from proof diffgetn-nodupssize (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 10:
n ≤ # x0, ∀ m0, n0. # x0 = m0 + 1 ∧ n0 ≤ m0 → #(x0 \ x0[n0] ') = m0 ∨ x0 = [] ⊦ 1 + #(x0 \ (a ' + x0)[n] ') = # x0, a = (a ' + x0)[n], x0 = [], dups(x0), a ∈ x0

Rule argument

fma: x0 = []

Additional information

Used simplifier rules: diff-e: ⊦ [] \ y = [];% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # [] = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ ¬ a ∈ []; (s)% ⊦ ¬ dups([]); (s)% ⊦ m + 1 ≠ 0; (s)% ⊦ a '[0] = a; (s) Used formulas: x0 = []