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Proof step 2 from proof ocs-del1-if (previous step)

⊦ #oc(a, x -1l b) = (a ≠ b ∨ ¬ b ∈ x ⊃ #oc(a, x);#oc(b, x) - 1)

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x);%del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1;% ⊦ n -1 = n - 1; (s)% ⊦ ¬ a ∈ x → x -1l a = x; (s)