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Proof step 3 from proof RED1-lem (previous step)

〈RED1(t1; tn)〉 tn = tn0, 〈TOPRED(t1 _ t2; tn)〉 tn = none, ∀ t0. #(t0) < #(t1 _ t2) → (true → 〈RED1(t0; tn)〉 (tn = none ⊃ irred(t0); t0 => tn .t)), ∀ t0. ¬ t1 _ t2 =>top t0, tn0 = none ⊃ irred(t1) ; t1 => tn0 .t, tn0 ≠ none ⊦ 〈tn0 := some(tn0 .t _ t2)〉 (tn0 = none ⊃ irred(t1 _ t2); t1 _ t2 => tn0 .t)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ some(t) ≠ none], [CSIMPSEQ : ⊦ some(t) .t = t], [CSIMPSEQ : ⊦ t => t' → (¬ t _ t1 => t' _ t1 ↔ false)])]]