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Proof step 21 from proof diff-sub-size (previous step)

#(s0 \ s) = # s0 - # s, a ∈ s0, a0 ∈ s0, # s0 < # s, a1 ∈ s, a2 ∈ s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s0 - # s - 1 = # s0 - # s + 1, a1 ∈ s0, a ∈ s

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 22:
#(s0 \ s) = # s0 - # s, a ∈ s0, a0 ∈ s0, # s0 < # s, a1 ∈ s, a2 ∈ s, a1 ∈ s ++ a → a1 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s0 - # s - 1 = # s0 - # s + 1, a1 ∈ s0, a ∈ s

Rule argument

pos: left 7 terms: a1

Additional information

Used terms: a1