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Proof step 3 from proof del1-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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%c: ⊦ a + x = a ' + x; from specification list