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Proof step 10 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) ⊦ 〈tn := some(t .1 .2)〉〈tn0 := tn ; if tn0 = none ∧ ap(t) then { RED1(t .1; tn0) ; if tn0 ≠ none then tn0 := some(tn0 .t _ t .2) else if value(t .1) then { RED1(t .2; tn0) ; if tn0 ≠ none then tn0 := some(t .1 _ tn0 .t) } }〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 9:
∀ 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) ⊦ 〈tn0 := some(t .1 .2)〉〈if tn0 = none ∧ ap(t) then { RED1(t .1; tn0) ; if tn0 ≠ none then tn0 := some(tn0 .t _ t .2) else if value(t .1) then { RED1(t .2; tn0) ; if tn0 ≠ none then tn0 := some(t .1 _ tn0 .t) } }〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]