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Proof step 24 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 tn = tn0, tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ap(ks(n0) _ K) ∧ ap((ks(n0) _ K) .1) ∧ (ks(n0) _ K) .1 .1 = K ∧ value((ks(n0) _ K) .1 .2) ∧ value((ks(n0) _ K) .2)), ap(ks(n0) _ K) ∧ ap((ks(n0) _ K) .1) ∧ ap((ks(n0) _ K) .1 .1) ∧ (ks(n0) _ K) .1 .1 .1 = S, v = (ks(n0) _ K) .2, v0 = (ks(n0) _ K) .1 .2, v4 = (ks(n0) _ K) .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v), tn = some(v4 _ v _ (v0 _ v)) ⊦ 〈skip〉 ((tn = none → n0 = 0) ∧ (tn ≠ none → n0 ≠ 0 ∧ tn = some(ks(n0 + 1 - 2))))

Heuristic symbolic execution applied rule skip right

And obtained the following new goals

Proof step 23:
〈RED1(ks(n0); tn)〉 tn = tn0, tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ap(ks(n0) _ K) ∧ ap((ks(n0) _ K) .1) ∧ (ks(n0) _ K) .1 .1 = K ∧ value((ks(n0) _ K) .1 .2) ∧ value((ks(n0) _ K) .2)), ap(ks(n0) _ K) ∧ ap((ks(n0) _ K) .1) ∧ ap((ks(n0) _ K) .1 .1) ∧ (ks(n0) _ K) .1 .1 .1 = S, v = (ks(n0) _ K) .2, v0 = (ks(n0) _ K) .1 .2, v4 = (ks(n0) _ K) .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v), tn = some(v4 _ v _ (v0 _ v)), ¬ (tn = none → n0 = 0) ∨ ¬ (tn ≠ none → n0 ≠ 0 ∧ tn = some(ks(n0 + 1 - 2))) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]