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Proof step 6 from proof delapp (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

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;% ⊦ n -1 = n - 1; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: a