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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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%filt-y: ⊦ a ∈ y → filter(a ' + x, y) = a ' + filter(x, y); from specification list-set