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Proof step 4 from proof subdel-03 (previous step)

x -l a ⊆m y -l a ⊦ #oc(a, x) ≤ #oc(a, y -1l a) ↔ #oc(a, x) < #oc(a, y) ∨ ¬ a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%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-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%in-02-01: ⊦ ¬ a ∈ x → x -1l a = x; from specification list-del%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%zero-01: ⊦ ¬ m < 0; from specification nat-basic1 Used terms: y, a