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Proof step 5 from proof termination-RED (previous step)

〈RED1(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), tn1 ≠ none, #(tn1) < #(t), ¬ hasS(tn1), ¬ hasS(t), tn1 = some(t1) ⊦ 〈RED(t1; t0)〉 true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
〈RED1(t; tn)〉 tn = some(t1), ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), #(t1) < #(t), ¬ hasS(t), ¬ hasS(t1) ⊦ 〈RED(t1; t0)〉 true

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ hasS(some(t)) ↔ hasS(t)], [CSIMPSEQ : ⊦ #(some(t)) = #(t)], [CSIMPSEQ : ⊦ some(t) ≠ none])]]