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Proof step 99 from proof termination-RED1 (previous step)

〈RED1(t2; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), value(t1), ¬ hasS(t1), ¬ hasS(t2), ¬ ap(t1), tn1 ≠ none → ¬ hasS(tn1) ∧ #(tn1) < #(t2), tn1 ≠ none, ¬ (some(t1 _ tn1 .t) ≠ none → ¬ hasS(some(t1 _ tn1 .t)) ∧ ¬ #(t1) + #(t2) < #(some(t1 _ tn1 .t))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 98:
〈RED1(t2; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), tn1 ≠ none, #(tn1) < #(t2), value(t1), ¬ hasS(tn1), ¬ ap(t1), ¬ hasS(t2), ¬ hasS(t1), ¬ (¬ hasS(tn1 .t) ∧ #(tn1 .t) < #(t2)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ m + n < m + n0 ↔ n < n0], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ #(t _ t0) = (#(t) + #(t0)) +1], [CSIMPSEQ : ⊦ #(some(t)) = #(t)], [CSIMPSEQ : ⊦ hasS(t _ t0) ↔ hasS(t) ∨ hasS(t0)], [CSIMPSEQ : ⊦ hasS(some(t)) ↔ hasS(t)], [CSIMPSEQ : ⊦ some(t) ≠ none])]]