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Proof step 18 from proof diff-sub-size (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 19:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 20:
∃ a. a ∈ s ∧ ¬ a ∈ s0, #(s0 \ s) = # s0 - # s, # s0 < # s, a0 ∈ s0, a ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ # s0 - # s - 1 = # s0 - # s + 1, a ∈ s

Rule argument

spec: set-basic inst: name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [s0, s] flags:

Additional information

Used terms: s0, s