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Proof step 12 from proof diff-sub-size (previous step)

a ∈ s0, a0 ∈ s0, (a1 ∈ s → a1 ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s

User applied rule case distinction

And obtained the following new goals

Proof step 13:
a1 ∈ s, a0 ∈ s0, a ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s, a1 ∈ s0
Proof step 15:
#(s0 \ s) = # s0 - # s, a0 ∈ s0, a ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s

Rule argument

pos: left 3

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