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Proof step 9 from proof termination-RED (previous step)

〈RED1(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), ¬ hasS(t), tn1 ≠ none → ¬ hasS(tn1) ∧ #(tn1) < #(t) ⊦ 〈if tn1 ≠ none then RED(tn1 .t; t0) else t0 := t〉 true

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 7:
〈RED1(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), ¬ hasS(t), tn1 ≠ none → ¬ hasS(tn1) ∧ #(tn1) < #(t), tn1 ≠ none ⊦ 〈RED(tn1 .t; t0)〉 true
Proof step 8:
〈RED1(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), ¬ hasS(t), tn1 ≠ none → ¬ hasS(tn1) ∧ #(tn1) < #(t), ¬ tn1 ≠ none ⊦ 〈t0 := t〉 true

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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