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Proof step 4 from proof size-three-01 (previous step)

# s < 3, c ∈ s, b ∈ s, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ a = b, a = c, b = c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b ∈ s0, c ∈ s0, # s0 < 2 ⊦ b = c, a ∈ s0

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s);%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-num-05: n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; from specification nat