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Proof step 59 from proof termination-RED1 (previous step)

〈if ap(t3 _ t4) ∧ ap((t3 _ t4) .1) ∧ (t3 _ t4) .1 .1 = K ∧ value((t3 _ t4) .1 .2) ∧ value((t3 _ t4) .2) then tn := some((t3 _ t4) .1 .2) else if ap(t3 _ t4) ∧ ap((t3 _ t4) .1) ∧ ap((t3 _ t4) .1 .1) ∧ (t3 _ t4) .1 .1 .1 = S then let v1 = (t3 _ t4) .1 .1 .2, v2 = (t3 _ t4) .1 .2, v3 = (t3 _ t4) .2 in if value(v1) ∧ value(v2) ∧ value(v3) then tn := some(v1 _ v3 _ (v2 _ v3)) else tn := none else tn := none〉 tn = none, 〈RED1(t4; tn2)〉 tn2 = none, 〈RED1(t3; tn1)〉 tn1 = none, ∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), value(t3 _ t4), value(t3), ¬ hasS(t2), ¬ hasS(t3), ¬ hasS(t4), ¬ (t3 = K ∧ value(t4) ∧ value(t2)), ¬ (ap(t3) ∧ t3 .1 = S) ⊦ 〈RED1(t2; tn0)〉 〈t1 := t3 _ t4 ; 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 : ()]]