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Proof step 58 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, tn = tn0 ⊦ 〈if tn ≠ none then tn := some(ks(n0) _ tn .t)〉 ((tn = none → n0 = 0) ∧ (tn ≠ none → n0 ≠ 0 ∧ tn = some(ks(n0 + 1 - 2))))

Heuristic symbolic execution applied rule if negative right

And obtained the following new goals

Proof step 57:
〈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, tn = tn0 ⊦ 〈skip〉 ((tn = none → n0 = 0) ∧ (tn ≠ none → n0 ≠ 0 ∧ tn = some(ks(n0 + 1 - 2))))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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