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Proof step 6 from proof links-dirs-union (previous step)

dirs0 ⊥ dirs1 ⊦ (alink.pino ∈ dirs0 ∨ alink.pino ∈ dirs1) ∧ alink.name ∈ (alink.pino ∈ dirs0 ⊃ dirs0[alink.pino];dirs1[alink.pino]).entries ∧ ino = (alink.pino ∈ dirs0 ⊃ dirs0[alink.pino];dirs1[alink.pino]).entries[alink.name] ↔ alink ∈ links(ino, dirs0) ∪ links(ino, dirs1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
alink = mklink(m, str), dirs0 ⊥ dirs1 ⊦ (m ∈ dirs0 ∨ m ∈ dirs1) ∧ str ∈ (m ∈ dirs0 ⊃ dirs0[m];dirs1[m]).entries ∧ ino = (m ∈ dirs0 ⊃ dirs0[m];dirs1[m]).entries[str] ↔ alink ∈ links(ino, dirs0) ∪ links(ino, dirs1)

Rule argument

spec: alink name: elim seq: ⊦ m = alink.pino ∧ str = alink.name ↔ alink = mklink(m, str) subst: [alink, m, str] → [alink, m, str]

Additional information

Used simplifier rules: ⊦ m = alink.pino ∧ str = alink.name ↔ alink = mklink(m, str); (elim) Used terms: alink, m, str