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Proof step 2 from proof del-size-02 (previous step)

a ∈ s ⊦ #(s -- b -- a) = #(s -- b) - 1, a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s, a0 ∈ s ⊦ #(s -- b -- a) = #(s -- b) - 1, a = b

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic