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Proof step 5 from proof oc-dups (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
n < #oc(a, a0 ') + #oc(a, x0), ∀ n0, a1. n0 < #oc(a1, x0) ∧ a1 ∈ x0 → #oc(a1, x0) = 1, a = a0 ∨ a ∈ x0 ⊦ a0 = a, dups(x0), a0 ∈ x0, a ≠ a0 ∧ #oc(a, x0) = 1, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < 1 ↔ m = 0; (s) Used formulas: a0 = a