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Proof step 32 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 tn = none, t0 = S, ap(ks(n0)), ap(ks(n0) .1), value(ks(n0)), n0 ≤ 1, ¬ (ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (value(t1) ∧ value(ks(n0) .2)), ks(n0) .1 = t0 _ t1 ⊦ 〈RED1(K; tn1)〉〈if tn1 ≠ none then tn1 := some(ks(n0) _ tn1 .t)〉 ((tn1 = none → n0 = 0) ∧ (tn1 ≠ none → n0 ≠ 0 ∧ tn1 = some(ks(n0 + 1 - 2))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
〈RED1(ks(n0); tn)〉 tn = none, ks(n0) .1 = S _ t1, n0 ≤ 1, value(ks(n0)), ap(ks(n0)), ¬ (value(t1) ∧ value(ks(n0) .2)) ⊦ 〈RED1(K; tn1)〉〈if tn1 ≠ none then tn1 := some(ks(n0) _ tn1 .t)〉 ((tn1 = none → n0 = 0) ∧ (tn1 ≠ none → n0 ≠ 0 ∧ tn1 = some(ks(n0 + 1 - 2))))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ t _ t0 ≠ K], [CSIMPSEQ : ⊦ ap(t _ t0)])]]