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Proof step 3 from proof ocs-del1 (previous step)

∀ a. a ∈ x0 → #oc(a, x0) = 1 ∨ a ∈ x0 -1l a, a ∈ a0 + x0 ⊦ #oc(a, a0 + x0) = 1, a ∈ (a0 + x0) -1l a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. a ∈ x0 → #oc(a, x0) = 1 ∨ a ∈ x0 -1l a, a = a0 ∨ a ∈ x0 ⊦ a ∈ (a0 ' + x0) -1l a, 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;%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