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Proof step 2 from proof sub-ade-03 (previous step)

s ⊆ s0 ⊦ a ∈ s, s ⊆ (s0 -- a) ++ b ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 ∈ s, a1 ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a0 = b, a0 = c, a0 ∈ s0, a ∈ s

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws)% ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)