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Proof step 8 from proof le (previous step)

# y < # filter(a ' + rmdup(x0), y), a ∈ y, ∀ y. # y < # filter(rmdup(x0), y) → filter(rmdup(x0), y) = [] ⊦ filter(a ' + rmdup(x0), y) = [], a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ y, # y ≤ # filter(rmdup(x0), y), ∀ y. # y < # filter(rmdup(x0), y) → filter(rmdup(x0), y) = [] ⊦ a ∈ x0

Rule argument

Additional information

Used simplifier rules: filt-y: ⊦ a ∈ y → filter(a ' + x, y) = a ' + filter(x, y);% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list