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Proof step 6 from proof diff-sub-size (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
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)
Proof step 29:
∀ s0. (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ a ∈ s0 \ s, #((s0 \ s) -- a) = # s0 - #(s ++ a)

Rule argument

fma: a ∈ s0 \ s

Additional information

Used formulas: a ∈ s0 \ s