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Proof step 2 from proof del-feq (previous step)

∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ a ≠ b ∧ b ∈ s ∧ ¬ a ∈ s0

User applied rule all left

And obtained the following new goals

Proof step 3:
b ≠ a ∧ b ∈ s ↔ b = b ∨ b ∈ s0, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ a ≠ b ∧ b ∈ s ∧ ¬ a ∈ s0

Rule argument

pos: left 1 terms: b abort: F discard: F computed: T

Additional information

Used terms: b