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Proof step 6 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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 7:
n ≤ # x0, ∀ m0, n0. # x0 = m0 + 1 ∧ n0 ≤ m0 → #(x0 \ x0[n0] ') = m0 ∨ x0 = [] ⊦ a ∈ (a ' + x0)[n] ', #((a ' + x0) \ (a ' + x0)[n] ') = # x0, dups(x0), a ∈ x0

Rule argument

fma: a ∈ (a ' + x0)[n] '

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%diff-onenonmember: ⊦ ¬ a ∈ x → x \ a ' = x;%a: ⊦ (x + y) + z = x + y + z; from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list Used formulas: a ∈ (a ' + x0)[n] '