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Proof step 7 from proof diffgetn-nodupssize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ≤ # x0, ∀ m0, n0. # x0 = m0 + 1 ∧ n0 ≤ m0 → #(x0 \ x0[n0] ') = m0 ∨ x0 = [] ⊦ #(x0 \ (a ' + x0)[n] ') + 1 = # 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);%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list