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Proof step 19 from proof termination-TOPRED (previous step)

¬ hasS(t) ⊦ 〈if ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2) then tn0 := 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 tn0 := some(v1 _ v3 _ (v2 _ v3)) else tn0 := none else tn0 := none〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 6:
¬ hasS(t), ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2) ⊦ 〈tn0 := some(t .1 .2)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))
Proof step 18:
¬ hasS(t), ¬ (ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)) ⊦ 〈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 tn0 := some(v1 _ v3 _ (v2 _ v3)) else tn0 := none else tn0 := none〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]