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Proof step 32 from proof diff-sub-size (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 33:
∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ a0 ∈ s → a0 ∈ s0, #(s0 \ s) = # s0 - #(s ++ a), a ∈ s0 \ s
Proof step 35:
#(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) = # s0 - #(s ++ a), a ∈ s0 \ s

Rule argument

pos: left 2

Additional information