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Proof step 5 from proof del1 (previous step)

#oc(a, x) = m + 1, a ∈ x ⊦ #oc(a, x -1l a) = m

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 6:
#oc(a, a0 + x0) = m + 1, a ∈ a0 + x0, ∀ a, m. #oc(a, x0) = m + 1 ∧ a ∈ x0 → #oc(a, x0 -1l a) = m ⊦ #oc(a, (a0 + x0) -1l a) = m

Rule argument

term: x

Additional information

Used simplifier rules: oc-e: ⊦ #oc(a, []) = 0;%list freely generated by [], +;;% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic