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

∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, ∀ a0. ¬ a0 ∈ s1 ++ a ∨ ¬ a0 ∈ s0 ⊦ #((s0 ∪ s1) ++ a) = #(s1 ++ a) + # s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ¬ a ∈ s → #(s ++ a) = # s + 1
Proof step 6:
¬ ¬ a ∈ s0 ∪ s1, ∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, ∀ a0. ¬ a0 ∈ s1 ++ a ∨ ¬ a0 ∈ s0 ⊦ #((s0 ∪ s1) ++ a) = #(s1 ++ a) + # s0
Proof step 14:
¬ a ∈ s0 ∪ s1, ∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, ∀ a0. ¬ a0 ∈ s1 ++ a ∨ ¬ a0 ∈ s0 ⊦ #(s0 ∪ s1) + 1 = #(s1 ++ a) + # s0

Rule argument

spec: set-basic inst: name: Size-insert seq: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1 subst: [a, s] → [a, s0 ∪ s1] flags: rewrite

Additional information

Used terms: a, s0 ∪ s1