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Proof step 7 from proof subseteq-rem-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ s0, s0 ⊆ s1, a ∈ s1, a0 = a ∨ a0 ∈ s1 \ s0 ⊦ s0 = ∅, # s1 = 0, s1 = ∅, # s0 = 0, 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;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);