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Proof step 10 from proof union-size-03 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
#(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #(s ∪ s0) + 1 = # s0 + # s + 1 - #(s0 ∩ s), a ∈ s, a ∈ s0, #(s0 ∩ s) ≤ # s0 + # s
Proof step 13:
#(s0 ∪ s) = n0, # s0 + # s = n0 + #(s0 ∩ s), #(s0 ∩ s) ≤ # s0 + # s ⊦ #(s ∪ s0) + 1 = # s0 + # s + 1 - #(s0 ∩ s), a ∈ s, a ∈ s0

Rule argument

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

Additional information

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