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Proof step 12 from proof union-eq-empty (previous step)

zh0 ⊥ zh1, ∀ r1. ¬ r1 ∈ zh0 ∧ ¬ r1 ∈ zh1 ⊦ ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

User applied rule all left

And obtained the following new goals

Proof step 13:
zh0 ⊥ zh1, ∀ r1. ¬ r1 ∈ zh0 ∧ ¬ r1 ∈ zh1 ⊦ r ∈ zh1, r ∈ zh0, ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

Rule argument

pos: left 2 terms: r abort: F discard: F computed: T

Additional information

Used terms: r