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Proof step 105 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), value(t1), ¬ hasS(t1), ¬ hasS(t2), ¬ ap(t1), none = none ⊦ 〈RED1(t2; tn0)〉〈if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 104:
∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), value(t1), ¬ hasS(t1), ¬ hasS(t2), ¬ ap(t1) ⊦ 〈RED1(t2; tn0)〉〈if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]