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Proof step 3 from proof btree-istore-reduce-refed (previous step)

((btree(rp, rp0, adr, height, dirty, ri, iast0, as, alladrs, nstat) * P))(zh), iast0 | as ⊆ iast1, as ⊆ dom iast0, as ⊆ alladrs ⊦ ((btree(rp, rp0, adr, height, dirty, ri, iast1, as, alladrs, nstat) * P))(zh)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
((btree(rp, rp0, adr, 0, dirty, ri, iast0, as, alladrs, nstat) * P))(zh), as ⊆ alladrs, as ⊆ dom iast0, iast0 | as ⊆ iast1 ⊦ ((btree(rp, rp0, adr, 0, dirty, ri, iast1, as, alladrs, nstat) * P))(zh)
Proof step 17:
((btree(rp, rp0, adr, height0 +1, dirty, ri, iast0, as, alladrs, nstat) * P))(zh), as ⊆ alladrs, as ⊆ dom iast0, iast0 | as ⊆ iast1, ∀ rp, rp0, adr, dirty, ri, iast0, as, alladrs, nstat, P, zh, iast1. ((btree(rp, rp0, adr, height0, dirty, ri, iast0, as, alladrs, nstat) * P))(zh) ∧ iast0 | as ⊆ iast1 ∧ as ⊆ dom iast0 ∧ as ⊆ alladrs → ((btree(rp, rp0, adr, height0, dirty, ri, iast1, as, alladrs, nstat) * P))(zh) ⊦ ((btree(rp, rp0, adr, height0 +1, dirty, ri, iast1, as, alladrs, nstat) * P))(zh)

Rule argument

indvar: height

Additional information

Used simplifier rules: nat freely generated by 0, +1;;