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Proof step 2 from proof del-size (previous step)

a ∈ s ⊦ s = ∅, #(s -- a) = # s - 1, # s = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-basic inst: name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s, a] flags: rewrite

Additional information

Used simplifier rules: del-addeq-04: ⊦ ¬ a ∈ s → (s ++ a) -- a = s;%ne: ⊦ ∅ ≠ s ++ a; from specification set-basic%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1; from specification set-basic%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: s, a