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Proof step 19 from proof subrec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
#oc(a0, y) < #oc(a0, x), a ∈ y, a0 ∈ x, ∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a) ⊦ a0 = a

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2