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Proof step 3 from proof app-03 (previous step)

(∀ a. #oc(a, y0) = 0) ↔ y0 = [] ⊦ (∀ a0. #oc(a0, a + y0) = 0) ↔ a + y0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
¬ a0 ∈ y0 → y0 = [], ∀ a1. #oc(a1, a ') + #oc(a1, y0) = 0 ⊦

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%oc-e: ⊦ #oc(a, []) = 0;%ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list