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Proof step 3 from proof oc-dups (previous step)

∀ n, a. n < #oc(a, x0) ∧ a ∈ x0 → #oc(a, x0) = 1 ∨ dups(x0), n < #oc(a, a0 + x0), a ∈ a0 + x0 ⊦ #oc(a, a0 + x0) = 1, dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < #oc(a, a0 ') + #oc(a, x0), a = a0 ∨ a ∈ x0, ∀ n0, a1. n0 < #oc(a1, x0) ∧ a1 ∈ x0 → #oc(a1, x0) = 1 ⊦ #oc(a, a0 ') + #oc(a, x0) = 0, a0 ∈ x0, dups(x0), a ≠ a0 ∧ #oc(a, x0) = 1, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list