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Proof step 123 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ¬ hasS(t), ¬ (ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)), ¬ (ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S), tn = none, tn = none ∧ ap(t) ⊦ 〈RED1(t .1; tn)〉〈if tn ≠ none then tn := some(tn .t _ t .2) else if value(t .1) then { RED1(t .2; tn) ; if tn ≠ none then tn := some(t .1 _ tn .t) }〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 122:
∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ap(t), ¬ hasS(t), ¬ (ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)), ¬ (ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S) ⊦ 〈RED1(t .1; tn)〉〈if tn ≠ none then tn := some(tn .t _ t .2) else if value(t .1) then { RED1(t .2; tn) ; if tn ≠ none then tn := some(t .1 _ tn .t) }〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]