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Proof step 8 from proof subseteq-rem-del (previous step)

a ∈ s1, a ∈ s0, s0 ⊆ s1 ⊦ # s1 = 0, # s0 = 0, a0 ∈ s1 ∧ ¬ a0 ∈ s0 -- a ↔ a0 = a ∨ a0 ∈ s1 \ s0, s0 = ∅, s1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ s0, a ∈ s1, s0 ⊆ s1, a0 = a ∨ a0 ∈ s1 \ s0 ⊦ # s0 = 0, # s1 = 0, s0 = ∅, s1 = ∅, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;