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Proof step 60 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 tn = none, value(ks(n0)), n0 ≤ 1, ¬ (ap(ks(n0)) ∧ ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (ap(ks(n0)) ∧ ap(ks(n0) .1) ∧ ks(n0) .1 .1 = S), tn0 = none ⊦ 〈skip〉〈tn1 := tn0 ; 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 symbolic execution applied rule skip right

And obtained the following new goals

Proof step 59:
〈RED1(ks(n0); tn)〉 tn = none, value(ks(n0)), n0 ≤ 1, ¬ (ap(ks(n0)) ∧ ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (ap(ks(n0)) ∧ ap(ks(n0) .1) ∧ ks(n0) .1 .1 = S), tn0 = none ⊦ 〈tn1 := tn0〉〈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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]