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Proof step 5 from proof oc-diff-01 (previous step)

a0 = a, ∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ #oc(a, (a0 ' + x0) \ y) = #oc(a, a0 ') + #oc(a, x0), a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ #oc(a, x0 \ y) = #oc(a, x0), a ∈ y

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%one: ⊦ #oc(a, a ') = 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y); from specification list-set