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Proof step 11 from proof del-02 (previous step)

#oc(a1, x) < #oc(a1, y), #oc(a1, y -1l a) ≤ #oc(a1, x -1l a), a ∈ y, a ∈ x, a1 ∈ y, ∀ a0. #oc(a0, x -1l a) ≤ #oc(a0, y -1l a) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: del1-01 seq: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x) subst: [b, x, a] → [a, y, a1] flags: rewrite, current

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);%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%pred-03: ⊦ n -1 = n - 1; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%ls-03: n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a, y, a1