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Proof step 36 from proof diff-sub-size (previous step)

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

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 37:
#(s0 \ s) = # s0 - # s, a ∈ s ++ a → a ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s0 - # s = # s0 - #(s ++ a), a ∈ s0 \ s

Rule argument

pos: left 2 terms: a

Additional information

Used terms: a