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Proof step 65 from proof termination-RED1 (previous step)

〈if tn3 ≠ none then tn3 := some(t1 .1 _ tn3 .t)〉 tn3 = none, 〈RED1(t1 .2; tn2)〉 tn2 = tn3, 〈RED1(t1 .1; tn1)〉 tn1 = tn2, 〈TOPRED(t1; tn0)〉 tn0 = tn1, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), value(t1), ¬ hasS(t1), ¬ hasS(t2), ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S), tn1 = none ∧ ap(t1), ¬ tn2 ≠ none, value(t1 .1) ⊦ 〈RED1(t2; tn0)〉 〈if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Heuristic conditional left split applied rule if left

And obtained the following new goals

Rule argument

[fmapos-arg: pos: left 1]

Additional information

[[CONCRETESIMPRULES : ()]]