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Proof step 129 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ¬ hasS(t) ⊦ 〈if ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2) then tn := some(t .1 .2) else if ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S then let v1 = t .1 .1 .2, v2 = t .1 .2, v3 = t .2 in if value(v1) ∧ value(v2) ∧ value(v3) then tn := some(v1 _ v3 _ (v2 _ v3)) else tn := none else tn := none〉 〈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 conditional right split applied rule if right

And obtained the following new goals

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]