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Proof step 3 from proof oc-del-02 (previous step)

#oc(a0, y -1l a) < #oc(a0, x -1l a), a0 ∈ x -1l a, a0 ∈ x, #oc(a0, x) ≤ #oc(a0, y) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: del1-01 seq: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x) subst: [b, x, a] → [a, y, a0] flags: <all paths>

Additional information

Used simplifier rules: del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x);% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ n -1 = n - 1; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%in-01-01: ⊦ a ∈ x -1l b → a ∈ x; (forward) from specification list-del% ⊦ ¬ n + (m - n) < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: a, y, a0