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Proof step 12 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 13:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 14:
# s = # s1 + 1, ∀ a0. a0 ∈ s → a0 ∈ s1, ∃ a. a ∈ s ∧ ¬ a ∈ s1 ⊦ a ∈ s1, 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: ⊦ m < n ↔ ¬ n ≤ m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: s1, s