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Proof step 4 from proof union-size-le (previous step)

# s0 + #(s ++ a) < #((s ∪ s0) ++ a), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ #((s ∪ s0) ++ a) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ s → s ++ a = s
Proof step 6:
¬ a ∈ s, # s0 + #(s ++ a) < #((s ∪ s0) ++ a), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ #((s ∪ s0) ++ a) = 0
Proof step 13:
a ∈ s, # s0 + # s < #((s ∪ s0) ++ a), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ #((s ∪ s0) ++ a) = 0

Rule argument

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

Additional information

Used terms: a, s