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Proof step 6 from proof ks-RED1 (previous step)

⊦ 〈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 assign right

And obtained the following new goals

Proof step 5:
⊦ 〈tn0 := none〉〈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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]