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Proof step 84 from proof termination-RED1 (previous step)

〈RED1(t3; tn1)〉 tn1 = 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), ¬ (ap(t3) ∧ t3 .1 = K ∧ value(t3 .2) ∧ value(t4)), ¬ (ap(t3) ∧ ap(t3 .1) ∧ t3 .1 .1 = S) ⊦ 〈RED1(t2; tn0)〉〈t1 := t3 _ t4 ; if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

User applied rule apply induction

And obtained the following new goals

Proof step 83:
〈RED1(t2; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t2)), 〈RED1(t3; tn1)〉 tn1 = 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), ¬ (ap(t3) ∧ t3 .1 = K ∧ value(t3 .2) ∧ value(t4)), ¬ (ap(t3) ∧ ap(t3 .1) ∧ t3 .1 .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

[substlist-arg: [t0] → [t2]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ #(t _ t0) = (#(t) + #(t0)) +1], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m + m0 ≤ m ↔ m0 = 0], [CSIMPSEQ : n < n0 ⊦ ((* n)) ≠ m0 + ((* n0))], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (t2)]]