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Proof step 3 from proof ocs-del-del1 (previous step)

#oc(a, a0 + x0) = 1, ∀ a. #oc(a, x0) = 1 → x0 -l a = x0 -1l a ⊦ (a0 + x0) -l a = (a0 + x0) -1l a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
#oc(a, a0 ') + #oc(a, x0) = 1, ∀ a. #oc(a, x0) = 1 → x0 -l a = x0 -1l a ⊦ (a0 ' + x0) -l a = (a0 ' + x0) -1l a

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)