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Proof step 24 from proof disj-reduce (previous step)

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))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ disj(T1, T2) ↔ (∀ n, m. n < # T1 ∧ m < # T2 ∧ T1[n] ≠ empty ∧ T2[m] ≠ empty → T1[n].elem ≠ T2[m].elem)

Rule argument

spec: inst: name:disj seq: ⊦ disj(T1, T2) ↔ (∀ n, m. n < # T1 ∧ m < # T2 ∧ T1[n] ≠ empty ∧ T2[m] ≠ empty → T1[n].elem ≠ T2[m].elem) subst:[T2, T1] → [T4, T3] flags:rewrite, current

Additional information

[Concretesimprules(()), Extraterms((T4T3))]