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Proof step 14 from proof subrec (previous step)

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1;%del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x);%one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%pred-03: ⊦ n -1 = n - 1; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2%sub-ls: ⊦ ¬ n + (m - n) < m; from specification nat Used formulas: a0 = a