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Proof step 4 from proof oc-diff-01 (previous step)

∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ #oc(a, (a0 ' + x0) \ y) = #oc(a, a0 ') + #oc(a, x0), a ∈ y

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a0 = a, ∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ #oc(a, (a0 ' + x0) \ y) = #oc(a, a0 ') + #oc(a, x0), a ∈ y
Proof step 7:
∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ a0 = a, #oc(a, (a0 ' + x0) \ y) = #oc(a, a0 ') + #oc(a, x0), a ∈ y

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a