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Proof step 27 from proof ks-TOPRED (previous step)

⊦ 〈if ap(ks(n)) ∧ ap(ks(n) .1) ∧ ks(n) .1 .1 = K ∧ value(ks(n) .1 .2) ∧ value(ks(n) .2) then tn0 := some(ks(n) .1 .2) else if ap(ks(n)) ∧ ap(ks(n) .1) ∧ ap(ks(n) .1 .1) ∧ ks(n) .1 .1 .1 = S then let v1 = ks(n) .1 .1 .2, v2 = ks(n) .1 .2, v3 = ks(n) .2 in if value(v1) ∧ value(v2) ∧ value(v3) then tn0 := some(v1 _ v3 _ (v2 _ v3)) else tn0 := none else tn0 := none〉 (tn0 ≠ none → n = 2 ∧ tn0 = some(K))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 14:
ap(ks(n)) ∧ ap(ks(n) .1) ∧ ks(n) .1 .1 = K ∧ value(ks(n) .1 .2) ∧ value(ks(n) .2) ⊦ 〈tn0 := some(ks(n) .1 .2)〉 (tn0 ≠ none → n = 2 ∧ tn0 = some(K))
Proof step 26:
¬ (ap(ks(n)) ∧ ap(ks(n) .1) ∧ ks(n) .1 .1 = K ∧ value(ks(n) .1 .2) ∧ value(ks(n) .2)) ⊦ 〈if ap(ks(n)) ∧ ap(ks(n) .1) ∧ ap(ks(n) .1 .1) ∧ ks(n) .1 .1 .1 = S then let v1 = ks(n) .1 .1 .2, v2 = ks(n) .1 .2, v3 = ks(n) .2 in if value(v1) ∧ value(v2) ∧ value(v3) then tn0 := some(v1 _ v3 _ (v2 _ v3)) else tn0 := none else tn0 := none〉 (tn0 ≠ none → n = 2 ∧ tn0 = some(K))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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