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Proof step 85 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 ((tn = none → n0 ≤ 1) ∧ (tn ≠ none → ¬ n0 ≤ 1 ∧ tn = some(ks(n0 - 2)))) ⊦ 〈RED1(ks(n0 +1); tn)〉 ((tn = none → n0 +1 ≤ 1) ∧ (tn ≠ none → ¬ n0 +1 ≤ 1 ∧ tn = some(ks(n0 +1 - 2))))

Heuristic split applied rule split left

And obtained the following new goals

Proof step 84:
〈RED1(ks(n0); tn)〉 tn = tn0, tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)) ⊦ 〈RED1(ks(n0 +1); tn)〉 ((tn = none → n0 +1 ≤ 1) ∧ (tn ≠ none → ¬ n0 +1 ≤ 1 ∧ tn = some(ks(n0 +1 - 2))))

Rule argument

[fmapos-arg: pos: left 1]

Additional information

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