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Proof step 54 from proof ks-RED (previous step)

∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))) ⊦ 〈if ap(K) ∧ ap(K .1) ∧ K .1 .1 = K ∧ value(K .1 .2) ∧ value(K .2) then tn := some(K .1 .2) else 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) } }〉〈n0 := 1 ; tn := tn0 ; if tn ≠ none then tn := some(ks(n0) .1 _ tn .t)〉〈if tn ≠ none then RED(tn .t; t0) else t0 := ks(n0)〉 ((even(n0) → t0 = K) ∧ (odd(n0) → t0 = K _ K))

Heuristic symbolic execution applied rule if negative right

And obtained the following new goals

Proof step 53:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))) ⊦ 〈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) } }〉〈n0 := 1 ; tn := tn0 ; if tn ≠ none then tn := some(ks(n0) .1 _ tn .t)〉〈if tn ≠ none then RED(tn .t; t0) else t0 := ks(n0)〉 ((even(n0) → t0 = K) ∧ (odd(n0) → t0 = K _ K))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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