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Proof step 2 from proof inv-isbad (previous step)

m < # pebs, mtd-cons(pebs) ⊦ ⟪mtd_asm_isbad#(m, pebs; ; isbad, err)⟫ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # pebs, mtd-cons(pebs) ⊦ ⟪mtd_asm_isbad#(m, pebs; ; isbad, err)⟫ true

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic