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Proof step 6 from proof union-size-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ s → s ++ a = s
Proof step 8:
¬ a ∈ s ∪ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #((s ∪ s0) ++ a) = # s0 + # s + 1 - #(s0 ∩ (s ++ a)), a ∈ s
Proof step 13:
a ∈ s ∪ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #(s ∪ s0) = # s0 + # s + 1 - #(s0 ∩ (s ++ a)), a ∈ s

Rule argument

spec: set-basic inst: name: eq seq: ⊦ a ∈ s → s ++ a = s subst: [a, s] → [a, s ∪ s0] flags: rewrite

Additional information

Used terms: a, s ∪ s0