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Proof step 11 from proof subsize-eq (previous step)

# s = # s1 + 1, ∀ a0. a0 ∈ s → a0 ∈ s1 ⊦ a ∈ s, a ∈ s1

User applied rule insert lemma

And obtained the following new goals

Proof step 12:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 13:
∃ a. a ∈ s ∧ ¬ a ∈ s1, # s = # s1 + 1, ∀ a0. a0 ∈ s → a0 ∈ s1 ⊦ a ∈ s, a ∈ s1

Rule argument

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

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq: ⊦ m = n + 1 → ¬ m ≤ n; from specification nat Used terms: s1, s