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Proof step 3 from proof ocs-not-more (previous step)

∀ a. ¬(# x0 < #oc(a, x0) ∧ a ∈ x0), #(a0 + x0) < #oc(a, a0 + x0), a ∈ a0 + x0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x0 + 1 < #oc(a, a0 ') + #oc(a, x0), a = a0 ∨ a ∈ x0, ∀ a1. ¬ # x0 < #oc(a1, x0) ∨ ¬ a1 ∈ x0 ⊦ #oc(a, a0 ') + #oc(a, x0) = 0

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list