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Proof step 3 from proof disj-neq (previous step)

T1[n] = full(a), T2[m] = full(a), m < # T2, n < # T1, disj(T1, T2) ⊦

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] → [T2, T1] flags:rewrite, current

Additional information

[Concretesimprules(()), Extraterms((T2T1))]