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Proof step 5 from proof size-three-01 (previous step)

b ∈ s0, c ∈ s0, # s0 < 2 ⊦ b = c, a ∈ s0

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] → [s0, b] flags: rewrite, current

Additional information

Used simplifier rules: in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s);%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%add: ⊦ ¬ a ∈ s ++ b ↔ a ≠ b ∧ ¬ a ∈ s;%size-zero-02: ⊦ # s = 0 → s = ∅;%ne-01: ⊦ a ∈ s → s ≠ ∅;%l-num-05: n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: s0, b