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Proof step 5 from proof union-size-02 (previous step)

a0 ∈ s0, a ∈ s0 ⊦ #(s ∪ s0) = (#(s ∪ s0) - 1) + 1, a ∈ s

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
a ∈ s0, a0 ∈ s0 ⊦ #(s ∪ s0) = (#(s ∪ s0) - 1) + 1, a ∈ s, 1 ≤ #(s ∪ s0)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [#(s ∪ s0), 1, n]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: #(s ∪ s0), 1, n