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Proof step 6 from proof sizesubl (previous step)

# s1 < # s, s ⊆ s1 ++ a ⊦ a ∈ s1, a ∈ s

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s

Rule argument

spec: <toplevel> name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [s1, s]

Additional information

Used simplifier rules: sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0;%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; Used terms: s1, s