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

a ∈ x0, ∀ m0, a1. # x0 = m0 + 1 ∧ a1 ∈ x0 → #(x0 \ a1 ') = m0 ∨ x0 = [] ⊦ a0 ∈ a ', #((a0 ' + x0) \ a ') = # x0, dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ x0, ∀ m0, a1. # x0 = m0 + 1 ∧ a1 ∈ x0 → #(x0 \ a1 ') = m0 ∨ x0 = [] ⊦ #(x0 \ a ') + 1 = # x0, dups(x0), a0 ∈ 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