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Proof step 51 from proof ks-RED1 (previous step)

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

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 50:
〈RED1(ks(n0); tn)〉 tn = tn1, tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ap(ks(n0)) ∧ ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (ap(ks(n0)) ∧ ap(ks(n0) .1) ∧ ks(n0) .1 .1 = S), tn1 = tn0, tn1 ≠ none, ¬ (some(tn1 .t _ K) = none → n0 = 0) ∨ ¬ (some(tn1 .t _ K) ≠ none → n0 ≠ 0 ∧ some(tn1 .t _ K) = some(ks(n0 + 1 - 2))) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]