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Proof step 41 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 tn = tn1, ks(n0) .1 .1 = S, ap(ks(n0)), ap(ks(n0) .1), tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (value(ks(n0) .1 .2) ∧ value(ks(n0) .2)), tn1 = tn0 ⊦ 〈if tn1 ≠ none then tn1 := some(tn1 .t _ K) else if value(ks(n0)) then { RED1(K; tn1) ; if tn1 ≠ none then tn1 := some(ks(n0) _ tn1 .t) }〉 ((tn1 = none → n0 = 0) ∧ (tn1 ≠ none → n0 ≠ 0 ∧ tn1 = some(ks(n0 + 1 - 2))))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 30:
〈RED1(ks(n0); tn)〉 tn = tn1, ks(n0) .1 .1 = S, ap(ks(n0)), ap(ks(n0) .1), tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (value(ks(n0) .1 .2) ∧ value(ks(n0) .2)), tn1 = tn0, tn1 ≠ none ⊦ 〈tn1 := some(tn1 .t _ K)〉 ((tn1 = none → n0 = 0) ∧ (tn1 ≠ none → n0 ≠ 0 ∧ tn1 = some(ks(n0 + 1 - 2))))
Proof step 40:
〈RED1(ks(n0); tn)〉 tn = tn1, ks(n0) .1 .1 = S, ap(ks(n0)), ap(ks(n0) .1), tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (value(ks(n0) .1 .2) ∧ value(ks(n0) .2)), tn1 = tn0, ¬ tn1 ≠ none ⊦ 〈if value(ks(n0)) then { RED1(K; tn1) ; if tn1 ≠ none then tn1 := some(ks(n0) _ tn1 .t) }〉 ((tn1 = none → n0 = 0) ∧ (tn1 ≠ none → n0 ≠ 0 ∧ tn1 = some(ks(n0 + 1 - 2))))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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