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Proof step 8 from proof diffgetn-nodupssize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
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

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)