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Proof step 7 from proof firstN-01 (previous step)

list2set(x) ∩ s = ∅, list2set(x) ∩ s0 = ∅, a ∈ x, a ∈ firstn(n, x), a ∈ s ∨ a ∈ s0, n < # x ⊦ s ∪ s0 = ∅, #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, x = [], #(s ∪ s0) = 0

User applied rule case distinction

And obtained the following new goals

Proof step 8:
a ∈ s, list2set(x) ∩ s = ∅, list2set(x) ∩ s0 = ∅, a ∈ x, a ∈ firstn(n, x), n < # x ⊦ s ∪ s0 = ∅, #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, x = [], #(s ∪ s0) = 0
Proof step 13:
a ∈ s0, list2set(x) ∩ s = ∅, list2set(x) ∩ s0 = ∅, a ∈ x, a ∈ firstn(n, x), n < # x ⊦ s ∪ s0 = ∅, #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, x = [], #(s ∪ s0) = 0

Rule argument

pos: left 5

Proof step 7 from proof firstN-01 (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information