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Proof step 1 from proof sub-sizeneqex (conclusion)

s ⊆ s0, # s ≠ # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s

Rule argument

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

Additional information

Used simplifier rules: sub-size: ⊦ s ⊆ s0 → ¬ # s0 < # s;%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1 Used terms: s, s0