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Proof step 16 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), tn = some((ks(n0) _ K) .1 .2) ⊦ 〈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 15:
〈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), tn = some((ks(n0) _ K) .1 .2), ¬ (tn = none → n0 = 0) ∨ ¬ (tn ≠ none → n0 ≠ 0 ∧ tn = some(ks(n0 + 1 - 2))) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]