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Proof step 117 from proof termination-RED1 (previous step)

〈RED1(t1; tn)〉 tn = tn0, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ¬ hasS(t1), ¬ hasS(t2), ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S), tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t1), ¬ tn0 ≠ none ⊦ 〈if value(t1) then { RED1(t2; tn0) ; if tn0 ≠ none then tn0 := some(t1 _ tn0 .t) }〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 115:
〈RED1(t1; tn)〉 tn = tn0, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ¬ hasS(t1), ¬ hasS(t2), ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S), tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t1), ¬ tn0 ≠ none, value(t1) ⊦ 〈RED1(t2; tn0)〉〈if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))
Proof step 116:
〈RED1(t1; tn)〉 tn = tn0, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ¬ hasS(t1), ¬ hasS(t2), ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S), tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t1), ¬ tn0 ≠ none, ¬ value(t1) ⊦ 〈skip〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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