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Proof step 7 from proof diff-sub-size (previous step)

a ∈ s0 \ s, ∀ s0. (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #((s0 \ s) -- a) = # s0 - #(s ++ a)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ s → #(s -- a) = # s - 1
Proof step 9:
a ∈ s0 \ s, a ∈ s0 \ s, ∀ s0. (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - #(s ++ a)

Rule argument

spec: set-union inst: name: del-size seq: ⊦ a ∈ s → #(s -- a) = # s - 1 subst: [a, s] → [a, s0 \ s] flags: rewrite, current

Additional information

Used terms: a, s0 \ s