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Proof step 34 from proof TOPRED-lem (previous step)

¬ (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 ⊃ ¬ (∃ t0. t =>top t0); t =>top tn0 .t)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 27:
¬ (ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)), ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S ⊦ 〈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〉 (tn0 = none ⊃ ¬ (∃ t0. t =>top t0); t =>top tn0 .t)
Proof step 33:
¬ (ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)), ¬ (ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S) ⊦ 〈tn0 := none〉 (tn0 = none ⊃ ¬ (∃ t0. t =>top t0); t =>top tn0 .t)

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]