6 Section subst1_subst1. (**************************************************)
8 Theorem subst1_subst1: (t1,t2,u2:?; j:?) (subst1 j u2 t1 t2) ->
9 (u1,u:?; i:?) (subst1 i u u1 u2) ->
10 (EX t | (subst1 j u1 t1 t) & (subst1 (S (plus i j)) u t t2)).
11 Intros until 1; XElim H; Clear t2.
12 (* case 1: subst1_refl on first premise *)
14 (* case 2: subst1_single on first premise *)
15 Intros until 2; InsertEq H0 u2; XElim H0; Clear y; Intros.
16 (* case 2.1: subst1_refl on second premise *)
17 Rewrite H0; Clear H0 u1; XEAuto.
18 (* case 2.2: subst1_single on second premise *)
19 Rewrite H1 in H0; Clear H1 t0; Subst0Subst0; XEAuto.
22 Theorem subst1_subst1_back: (t1,t2,u2:?; j:?) (subst1 j u2 t1 t2) ->
23 (u1,u:?; i:?) (subst1 i u u2 u1) ->
24 (EX t | (subst1 j u1 t1 t) & (subst1 (S (plus i j)) u t2 t)).
25 Intros until 1; XElim H; Clear t2.
26 (* case 1: subst1_refl on first premise *)
28 (* case 2: subst1_single on first premise *)
29 Intros until 2; XElim H0; Clear u1; Intros;
30 Try Subst0Subst0; XEAuto.
33 Theorem subst1_trans: (t2,t1,v:?; i:?) (subst1 i v t1 t2) ->
34 (t3:?) (subst1 i v t2 t3) ->
36 Intros until 1; XElim H; Clear t2.
37 (* case 1: subst1_refl on first premise *)
39 (* case 2: subst1_single on first premise *)
40 Intros until 2; XElim H0; Clear t3; XEAuto.
45 Hints Resolve subst1_trans : ltlc.
47 Tactic Definition Subst1Subst1 :=
49 | [ H1: (subst1 ?0 ?1 ?2 ?3); H2: (subst1 ?4 ?5 ?6 ?1) |- ? ] ->
50 LApply (subst1_subst1 ?2 ?3 ?1 ?0); [ Intros H_x | XAuto ];
51 LApply (H_x ?6 ?5 ?4); [ Clear H_x; Intros H_x | XAuto ];
53 | [ H1: (subst1 ?0 ?1 ?2 ?3); H2: (subst0 ?4 ?5 ?6 ?1) |- ? ] ->
54 LApply (subst1_subst1 ?2 ?3 ?1 ?0); [ Intros H_x | XAuto ];
55 LApply (H_x ?6 ?5 ?4); [ Clear H_x; Intros H_x | XAuto ];
57 | [ H1: (subst1 ?0 ?1 ?2 ?3); H2: (subst1 ?4 ?5 ?1 ?6) |- ? ] ->
58 LApply (subst1_subst1_back ?2 ?3 ?1 ?0); [ Intros H_x | XAuto ];
59 LApply (H_x ?6 ?5 ?4); [ Clear H_x; Intros H_x | XAuto ];