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Proof step 25 from proof disj-reduce (conclusion)

⊦ disj(T1, T2) → ( ¬ disj(T3, T4) ↔ ¬ (∀ n, m. (n < # T1 ∧ m < # T2 ∧ T1[n] ≠ empty ∧ T2[m] ≠ empty → T1[n].elem ≠ T2[m].elem) → n < # T3 ∧ m < # T4 ∧ T3[n] ≠ empty ∧ T4[m] ≠ empty → T3[n].elem ≠ T4[m].elem))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
disj(T1, T2) ⊦ ¬ disj(T3, T4) ↔ (∃ n, m. ¬ ( (n < # T1 ∧ m < # T2 ∧ T1[n] ≠ empty ∧ T2[m] ≠ empty → T1[n].elem ≠ T2[m].elem) → n < # T3 ∧ m < # T4 ∧ T3[n] ≠ empty ∧ T4[m] ≠ empty → T3[n].elem ≠ T4[m].elem))

Rule argument

Additional information

[Concretesimprules(())]