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Proof step 6 from proof perm-switch (previous step)

n < # y, m < # y ⊦ (∀ a. a ∈ x ∨ a ∈ y[m, y[n]][n, y[m]] → #oc(a, x) = #oc(a, y[m, y[n]][n, y[m]])) ↔ (∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y)), y = []

User applied rule case distinction

And obtained the following new goals

Proof step 7:
∀ a. a ∈ x ∨ a ∈ y[m, y[n]][n, y[m]] → #oc(a, x) = #oc(a, y[m, y[n]][n, y[m]]), n < # y, m < # y ⊦ ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), y = []
Proof step 9:
∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), n < # y, m < # y ⊦ ∀ a. a ∈ x ∨ a ∈ y[m, y[n]][n, y[m]] → #oc(a, x) = #oc(a, y[m, y[n]][n, y[m]]), y = []

Rule argument

pos: right 1

Proof step 6 from proof perm-switch (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information