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Proof step 7 from proof ks-RED1 (previous step)

⊦ 〈if ap(K) ∧ ap(K .1) ∧ ap(K .1 .1) ∧ K .1 .1 .1 = S then let v1 = K .1 .1 .2, v2 = K .1 .2, v3 = K .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(K) then { RED1(K .1; tn0) ; if tn0 ≠ none then tn0 := some(tn0 .t _ K .2) else if value(K .1) then { RED1(K .2; tn0) ; if tn0 ≠ none then tn0 := some(K .1 _ tn0 .t) } }〉 tn0 = none

Heuristic symbolic execution applied rule if negative right

And obtained the following new goals

Proof step 6:
⊦ 〈tn := none〉〈tn0 := tn ; if tn0 = none ∧ ap(K) then { RED1(K .1; tn0) ; if tn0 ≠ none then tn0 := some(tn0 .t _ K .2) else if value(K .1) then { RED1(K .2; tn0) ; if tn0 ≠ none then tn0 := some(K .1 _ tn0 .t) } }〉 tn0 = none

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ ap(K)])]]