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Proof step 30 from proof RED1-lem (previous step)

〈TOPRED(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (true → 〈RED1(t0; tn)〉 (tn = none ⊃ irred(t0); t0 => tn .t)), tn1 = none ⊃ (∀ t0. ¬ t =>top t0) ; t =>top tn1 .t, tn1 = none ∧ ap(t) ⊦ 〈RED1(t .1; tn1)〉〈if tn1 ≠ none then tn1 := some(tn1 .t _ t .2) else if value(t .1) then { RED1(t .2; tn1) ; if tn1 ≠ none then tn1 := some(t .1 _ tn1 .t) }〉 (tn1 = none ⊃ irred(t); t => tn1 .t)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
〈TOPRED(t; tn)〉 tn = none, ∀ t0. #(t0) < #(t) → (true → 〈RED1(t0; tn)〉 (tn = none ⊃ irred(t0); t0 => tn .t)), ap(t), ∀ t0. ¬ t =>top t0 ⊦ 〈RED1(t .1; tn1)〉〈if tn1 ≠ none then tn1 := some(tn1 .t _ t .2) else if value(t .1) then { RED1(t .2; tn1) ; if tn1 ≠ none then tn1 := some(t .1 _ tn1 .t) }〉 (tn1 = none ⊃ irred(t); t => tn1 .t)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]