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Proof step 7 from proof delete-shift (previous step)

a1 ≠ a ∧ a1 ∈ s ↔ a1 ∈ s0, a ≠ a ∧ a ∈ s ↔ a ∈ s0, a ∈ s, a0 ∈ s, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0 ⊦ (a1 ∈ s ↔ a1 ∈ s0 ++ a) ∧ ¬ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a1 ≠ a ∧ a1 ∈ s ↔ a1 ∈ s0, a0 ∈ s, a ∈ s, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 ∈ s0 ⊦ a1 ∈ s ↔ a1 = a ∨ a1 ∈ s0, a ∈ s0

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)