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Proof step 10 from proof RED1-lem (previous step)

〈RED1(t2; tn)〉 tn = tn1, 〈RED1(t1; tn)〉 tn = none, 〈TOPRED(t1 _ t2; tn)〉 tn = none, ∀ t0. #(t0) < #(t1 _ t2) → (true → 〈RED1(t0; tn)〉 (tn = none ⊃ irred(t0); t0 => tn .t)), irred(t1), value(t1), ∀ t0. ¬ t1 _ t2 =>top t0, tn1 = none ⊃ irred(t2) ; t2 => tn1 .t, ¬ tn1 ≠ none, tn1 = none ⊃ ¬ irred(t1 _ t2) ; ¬ t1 _ t2 => tn1 .t ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
〈RED1(t2; tn)〉 tn = none, 〈RED1(t1; tn)〉 tn = none, 〈TOPRED(t1 _ t2; tn)〉 tn = none, ∀ t0. #(t0) < #(t1 _ t2) → (true → 〈RED1(t0; tn)〉 (tn = none ⊃ irred(t0); t0 => tn .t)), irred(t1), value(t1), irred(t2), ¬ irred(t1 _ t2), ∀ t0. ¬ t1 _ t2 =>top t0 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]