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Proof step 102 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) ⊦ 〈if tn1 ≠ none then tn1 := some(t1 _ tn1 .t)〉 (tn1 ≠ none → ¬ hasS(tn1) ∧ ¬ #(t1) + #(t2) < #(tn1))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 100:
〈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 ⊦ 〈tn1 := some(t1 _ tn1 .t)〉 (tn1 ≠ none → ¬ hasS(tn1) ∧ ¬ #(t1) + #(t2) < #(tn1))
Proof step 101:
〈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 ⊦ 〈skip〉 (tn1 ≠ none → ¬ hasS(tn1) ∧ ¬ #(t1) + #(t2) < #(tn1))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]