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Proof step 1 from proof value-03 (previous step)

⊦ t1 ≠ K ∧ t1 ≠ S → (value(t1 _ t2) ↔ ¬ ¬ (ap(t1) ∧ t1 .1 = S ∧ value(t2) ∧ value(t1 .2)))

Lemma val-other-ap