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Proof step 90 from proof termination-RED1 (previous step)

〈RED1(t3; tn1)〉 tn1 = none, 〈TOPRED(t1; tn0)〉 tn0 = none, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), value(t1), ap(t1), ¬ hasS(t1), ¬ hasS(t2), ¬ value(t3), ¬ (t3 = K ∧ value(t4) ∧ value(t2)), ¬ (ap(t3) ∧ t3 .1 = S), t1 = t3 _ t4 ⊦ 〈RED1(t2; tn0)〉〈if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 89:
〈RED1(t3; tn1)〉 tn1 = none, 〈TOPRED(t3 _ t4; tn0)〉 tn0 = none, ∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t3 ≠ K, value(t3 _ t4), ¬ hasS(t2), ¬ value(t3), ¬ hasS(t3), ¬ hasS(t4), ¬ (ap(t3) ∧ t3 .1 = S) ⊦ 〈RED1(t2; tn0)〉〈t1 := t3 _ t4 ; if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ value(K)], [CSIMPSEQ : ⊦ ¬ hasS(K)], [CSIMPSEQ : ⊦ value(K _ t) ↔ value(t)], [CSIMPSEQ : ⊦ hasS(t _ t0) ↔ hasS(t) ∨ hasS(t0)], [CSIMPSEQ : ⊦ ap(t _ t0)])]]