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Proof step 14 from proof union-size-03 (previous step)

a ∈ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ s0 = ∅, #(s ∪ s0) = # s0 + # s + 1 - #(s0 ∩ (s ++ a)), # s0 = 0, a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ a ∈ s → s ∩ (s0 ++ a) = (s ∩ s0) ++ a
Proof step 16:
a ∈ s0, a ∈ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ s0 = ∅, #(s ∪ s0) = # s0 + # s + 1 - #((s0 ∩ s) ++ a), # s0 = 0, a ∈ s

Rule argument

spec: set-union inst: name: intersection-shift seq: ⊦ a ∈ s → s ∩ (s0 ++ a) = (s ∩ s0) ++ a subst: [a, s0, s] → [a, s, s0] flags: rewrite, current

Additional information

Used terms: a, s, s0