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Proof step 17 from proof subrec (previous step)

a0 = a, a ∈ y, #oc(a0, y) < #oc(a0, a ') + #oc(a0, x), ∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a) ⊦ #oc(a0, a ') + #oc(a0, x) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a ∈ y, #oc(a, y) ≤ #oc(a, x), ∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a) ⊦

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2